PHYS 202 - GENERAL PHYSICS II Kinematics in Two Dimensions Question Bank Set 2

PHYS 202 - GENERAL PHYSICS II -

Kinematics in Two Dimensions

Question Bank - Set 2

Liberty University

Question 1

Question

A stone is thrown from the top of a cliff with an initial velocity of 20 m/s at

an angle of 30◦above the horizontal. The stone lands on the ground 4 seconds

later. Calculate the height of the cliff.

Solution

Step 1: Break the initial velocity into its horizontal (vi,x) and vertical (vi,y)

components. The initial velocity can be broken down into its horizontal and

vertical components as follows: vi,x =vicos(θ)vi,y =visin(θ)

Given: vi= 20 m/s θ= 30◦

Substitute these values: vi,x = 20 m/s cos(30◦) = 20 m/s ·√3

2= 10√3 m/s

vi,y = 20 m/s sin(30◦) = 20 m/s ·1

2= 10 m/s

Therefore, the initial horizontal velocity is 10√3 m/s and the initial vertical

velocity is 10 m/s.

Step 2: Determine the time of flight. The time of flight can be determined

from the vertical motion of the stone. Using the formula: y=vi,yt+1

2ayt2

where: y= 0 (final vertical position) vi,y = 10 m/s (initial vertical velocity)

ay=−9.81 m/s2(acceleration due to gravity)

Substitute the given values and solve for t: 0 = 10t−1

2·9.81t24.905t2−10t=

0t(4.905t−10) = 0 t= 0 or t=10

4.905 ≈2.04 s

Since time cannot be negative, the time of flight is approximately 2.04 sec-

onds.

Step 3: Calculate the maximum height of the stone. The maximum height

can be determined using the vertical motion of the stone. The stone reaches its

maximum height when its vertical velocity becomes zero.

Using the kinematic equation: vf=vi,y +aytwhere: vf= 0 (final vertical

velocity) vi,y = 10 m/s (initial vertical velocity) ay=−9.81 m/s2(acceleration

due to gravity) t= 2.04 s (time of flight)

Substitute the given values: 0 = 10−9.81 ×2.04 0 = 10 −19.9724 19.9724 =

The stone cannot reach such a height. Let’s recalculate with the correct

formula.

The height hof the cliff can be calculated using the horizontal motion of the

stone and the time of flight. The horizontal distance traveled by the stone is:

x=vi,xt

Given: vi,x = 10√3 m/s t= 4 s

Substitute these values: x= 10√3×4 = 40√3 m

Since the stone falls from the top of the cliff to the ground, the height of

the cliff is equal to the vertical position at t= 0. Using the formula: y=

vi,yt+1

2ayt2where: y=h(height of the cliff) vi,y = 10 m/s (initial vertical

velocity) ay=−9.81 m/s2(acceleration due to gravity) t= 4 s

Substitute the given values and solve for

Question 2

Question

A projectile is launched from the ground with an initial speed of 50 m/s at an

angle of 30◦above the horizontal. Find the total time the projectile is in the

air before hitting the ground.

Solution

Step 1: Resolve the initial velocity vector into its horizontal and vertical compo-

nents. The initial velocity vector v0can be expressed as the sum of its horizontal

component v0xand its vertical component v0y:

v0x=v0cos θ= 50 ×cos(30◦) = 43.3 m/s

v0y=v0sin θ= 50 ×sin(30◦) = 25 m/s

Step 2: Calculate the time of flight. The time of flight can be found using

the vertical motion equation:

y=v0yt+1

2ayt2

where yis the vertical displacement (final height - initial height) and ayis the

acceleration due to gravity (-9.81 m/s2). Substitute y= 0 (since the projectile

returns to the initial height) and solve for t:

0 = 25t−4.91t2

4.91t2−25t= 0

t(4.91t−25) = 0

t= 0 s or t=25

4.91 ≈5.09 s

Step 3: Double the time of flight to find the total time in the air. Since the

projectile is in the air for the same duration both upward and downward, the

total time in the air is:

Total time = 2 ×5.09 s = 10.18 s

Therefore, the total time the projectile is in the air before hitting the ground

is approximately 10.18 seconds.

Question 3

Question

A cannonball is fired at an angle of 30◦above the horizontal from the top of

a cliff that is 80 meters high. If the initial speed of the cannonball is 40 m/s,

determine the time it takes for the cannonball to hit the ground. Ignore air

resistance and assume g= 9.8 m/s2.

Solution

Step 1: Resolve the initial velocity of the cannonball into horizontal and vertical

components. Let v0= 40 m/s be the initial speed, θ= 30◦be the angle above the

horizontal, v0xbe the horizontal component, and v0ybe the vertical component.

v0x=v0cos(θ)

= 40 m/s ·cos(30◦)

= 40 m/s ·√3

= 20√3 m/s

v0y=v0sin(θ) = 40 m/s ·sin(30◦) = 40 m/s ·1

2= 20 m/s

Step 2: Determine the time it takes for the cannonball to reach the ground

using the vertical component. The vertical equation of motion is given by:

y=v0yt+1

2gt2

Since the cannonball starts at a height of 80 meters above the ground, when

it hits the ground, y=−80 meters. Thus, the vertical equation becomes:

−80 = 20t−1

2·9.8·t2

−80 = 20t−4.9t2

Solving for t:

4.9t2−20t−80 = 0

Using the quadratic formula, we find:

t=−(−20) ±p(−20)2−4·4.9·(−80)

2·4.9

t=20 ±√400 + 1568

9.8

t=20 ±√1968

9.8

t≈20 ±44.36

9.8

which gives us:

t1≈20 + 44.36

9.8=64.36

9.8≈6.57 s

t2≈20 −44.36

9.8=−24.36

9.8=−2.49 s

Since time cannot be negative, the time it takes for the cannonball to hit

the ground is approximately 6.57 seconds.

Question 4

Question

A particle is projected from the origin with an initial velocity of 30 m/s at an

angle of 30◦above the horizontal. Calculate the time when the particle is at

a height of 15 m above the origin and determine the horizontal distance it has

traveled at that time.

Solution

Let’s first analyze the motion of the particle in the vertical direction and then

in the horizontal direction.

Vertical Direction:



Given: v0y= 30 m/s ·sin(30◦) = 15 m/s, ay=−9.8 m/s2, and yf= 15 m.



We will use the kinematic equation yf=y0+v0yt+1

2ayt2to solve for the

time t.



Substituting the given values, we have 15 = 0 + 15t−1

2·9.8t2.



This equation simplifies to 4.9t2−15t+ 15 = 0.



We can solve this quadratic equation to find the positive value of t.

Horizontal Direction:



Given: v0x= 30 m/s ·cos(30◦) = 26 m/s.



The horizontal distance xtraveled by the particle at time tis given by

x=v0xt.



We will substitute v0xand the calculated value of tto find x.

Question 5

Question

A baseball player throws a ball with an initial velocity of 20 m/s at an angle of

30 degrees above the horizontal. The ball lands on the ground 100 meters away

from the player. Find the maximum height reached by the ball during its flight.

Solution

Step 1: Resolve the initial velocity of the ball into horizontal and vertical com-

ponents. The horizontal component of the initial velocity (v0x) can be found

using:

v0x=v0cos(θ)

where v0= 20 m/s and θ= 30◦.

v0x= 20 m/s ×cos(30◦)

v0x= 20 m/s ×√3

v0x= 10√3 m/s

The vertical component of the initial velocity (v0y) can be found using:

v0y=v0sin(θ)

v0y= 20 m/s ×sin(30◦)

v0y= 20 m/s ×1

v0y= 10 m/s

Step 2: Find the time taken to reach the peak height. Since the vertical

motion of the ball is symmetrical, the time taken to reach the peak height is

half of the total time of flight. The equation for the vertical motion of the ball

is:

y=v0yt−1

2gt2

where yis the maximum height reached. At the peak height, the vertical velocity

becomes 0.

0 = v0y−gt

t=v0y

t=10 m/s

9.81 m/s2

t≈1.02 s

Step 3: Find the maximum height reached by the ball. Substitute the value

of tinto the equation for vertical motion:

y= 10 m/s ×1.02 s −1

2×9.81 m/s2×(1.02 s)2

y= 10.2 m −1

2×9.81 m/s2×1.04 s2

y= 10.2 m −1

2×9.81 m/s2×1.04 s2

y= 10.2 m −5.088 m

y≈5.112 m

Therefore, the maximum height reached by the ball during its flight is ap-

proximately 5.112 meters.

Question 6

Question

A baseball is hit with an initial velocity of 20 m/s at an angle of 30◦above the

horizontal. Ignoring air resistance, calculate the time it takes for the baseball

to reach its maximum height.

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The horizontal component is given by:

v0x=v0cos(θ) = 20 cos(30◦)≈17.32 m/s

The vertical component is given by:

v0y=v0sin(θ) = 20 sin(30◦)≈10 m/s

Step 2: Determine the time it takes to reach maximum height.

The baseball reaches maximum height when the vertical component of its

velocity is 0 m/s. We can use the equation vy=v0y−gt, where vyis the vertical

component of velocity at a given time t,v0yis the initial vertical component of

velocity, gis the acceleration due to gravity (9.8 m/s2), and tis the time taken.

Setting vy= 0 and v0y= 10 m/s, we get:

0 = 10 −9.8t

t=10

9.8≈1.02 s

Therefore, it takes approximately 1.02 seconds for the baseball to reach its

maximum height.

Question 7

Question

A football quarterback throws a pass to a stationary receiver who is 20 meters

away. The initial velocity of the football is 15 m/s at an angle of 30◦above the

horizontal. What is the maximum height above the quarterback’s hand that

the ball reaches during its flight?

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

Let v0= 15 m/s be the initial velocity and θ= 30◦be the angle above the hor-

izontal. The horizontal component is given by v0x=v0cos(θ) and the vertical

component is given by v0y=v0sin(θ). Therefore, v0x= 15 cos(30◦) = 13.0 m/s

and v0y= 15 sin(30◦)=7.5 m/s.

Step 2: Determine the time of flight. Using the vertical component of ve-

locity, we can find the time it takes for the ball to reach its maximum height

using the equation vf=vi+at, where a=−9.8 m/s2is the acceleration due to

gravity and vf= 0 m/s at the maximum height. So, 0 = 7.5−9.8twhich gives

t= 0.77 s.

Step 3: Calculate the maximum height. The maximum height hcan be

found using the equation h=v0yt+1

2at2. Substitute v0y= 7.5 m/s, t= 0.77 s,

and a=−9.8 m/s2to get: h= 7.5(0.77) + 1

2(−9.8)(0.77)2= 5.77 m.

Therefore, the maximum height above the quarterback’s hand that the ball

reaches during its flight is 5.77 meters.

Question 8

Question

A baseball is hit at an angle of 30◦above the horizontal with an initial speed of

25 m/s. What are the horizontal and vertical components of its velocity at the

highest point of its trajectory?

Solution

Step 1: Identify the given information.

The initial speed of the baseball is v0= 25 m/s and the launch angle is θ= 30◦.

We will need to find the horizontal and vertical components of the velocity at

the highest point of its trajectory.

Step 2: Break the initial velocity into horizontal and vertical components.

The initial velocity can be broken down into its horizontal and vertical compo-

nents using trigonometric functions. The horizontal component can be found

using v0x=v0cos(θ) where θ= 30◦:

v0x= 25 m/s ·cos(30◦)

v0x= 25 ·√3

2≈21.65 m/s

The vertical component can be found using v0y=v0sin(θ) where θ= 30◦:

v0y= 25 m/s ·sin(30◦)

v0y= 25 ·1

2= 12.5 m/s

Step 3: Determine the horizontal and vertical components of velocity at the

highest point.

At the highest point of the trajectory, the vertical component of the velocity

is momentarily zero. Therefore, the horizontal component remains constant at

v0xand the vertical component becomes 0 at the highest point.

Thus, at the highest point of the trajectory: Horizontal component (vhx)

= 21.65 m/s Vertical component (vhy) = 0

Question 9

Question

A projectile is launched with an initial speed of 30 m/s at an angle of 60◦above

the horizontal. Find the time to reach maximum height, the maximum height

above the launch point, and the total time in the air.

Solution

Step 1: Break the initial velocity into its horizontal and vertical components.

The initial velocity of the projectile can be broken down into its horizontal and

vertical components as follows:

vix=vicos θ= 30 cos 60◦= 15 m/s

viy=visin θ= 30 sin 60◦= 25√3 m/s

Step 2: Find the time to reach maximum height. The time to reach maxi-

mum height can be found using the equation vf=vi+at, where vf= 0 at the

maximum height and a=−9.8 m/s2is the acceleration due to gravity. Solving

for tyields:

0 = 25√3−9.8t

t=25√3

9.8≈5.03 s

Step 3: Find the maximum height above the launch point. The maximum

height above the launch point can be found using the kinematic equation ymax =

yi+viyt+1

2at2, where yi= 0 and a=−9.8 m/s2. Thus, we have:

ymax = 25√3×5.03 −0.5×9.8×(5.03)2≈63.45 m

Step 4: Find the total time in the air. The total time in the air can be found

by doubling the time it took to reach the maximum height. Thus,

Total time = 2 ×5.03 = 10.06 s

Therefore, the projectile will reach its maximum height after approximately

5.03 seconds, at a height of 63.45 meters, and will be in the air for a total of

approximately 10.06 seconds.

Question 10

Question

A projectile is launched from the ground with an initial velocity of 20 m/s at

an angle of 30◦above the horizontal. Find the time it takes for the projectile

to reach its maximum height.

Solution

Step 1: Break the initial velocity into its horizontal and vertical components.

The horizontal component of the initial velocity is given by:

V0x=V0cos(θ)

where V0xis the horizontal component of the initial velocity, V0is the initial

velocity, and θis the launch angle. Substitute V0= 20 m/s and θ= 30◦:

V0x= 20 m/s ·cos(30◦)

V0x= 20 m/s ·√3

V0x= 10√3 m/s

The vertical component of the initial velocity is given by:

V0y=V0sin(θ)

where V0yis the vertical component of the initial velocity. Substitute V0=

20 m/s and θ= 30◦:

V0y= 20 m/s ·sin(30◦)

V0y= 20 m/s ·1

V0y= 10 m/s

Step 2: Determine the time to reach maximum height. At the maximum

height, the vertical component of the velocity becomes zero. Using the kinematic

equation for vertical motion:

Vy=V0y−gt

where Vyis the vertical component of the velocity, gis the acceleration due to

gravity (9.8 m/s2), and tis the time. Substitute Vy= 0, V0y= 10 m/s, and

g= 9.8 m/s2:

0 = 10 −9.8t

t=10

9.8

t≈1.02 s

Therefore, the time it takes for the projectile to reach its maximum height

is approximately 1.02 seconds.

Question 11

Question

A projectile is launched from the ground at an angle of 30◦above the horizontal

with an initial speed of 20 m/s. At its highest point, the projectile explodes into

two fragments of equal mass. One fragment continues in the original direction

with a speed of 8 m/s. What is the speed of the other fragment just after the

explosion?

Solution

Step 1: Analyze the projectile motion before the explosion. Let’s denote the

initial speed of the projectile as v0= 20 m/s. The horizontal component of

the initial velocity is v0x=v0cos(30◦) and the vertical component is v0y=

v0sin(30◦).

Step 2: Find the time to reach the highest point. The time it takes for the

projectile to reach its highest point can be found using the vertical component

of the motion. The vertical component of the velocity at the highest point is

zero. We can use the equation vy=v0y−gt where vy= 0 to find the time t.

0 = v0y−gt =⇒t=v0y

Step 3: Calculate the maximum height. The maximum height Hreached by

the projectile can be found using the equation H=v0y·t−1

2gt2.

H=v0y·v0y

g−1

2gv0y

g2

Step 4: Determine the horizontal distance traveled. The horizontal distance

Rtraveled by the projectile before the explosion can be found using the equation

R=vx·(2t) since the projectile will reach the same height on its way down as

it did on its way up.

R=v0x·(2t)

Step 5: Apply the conservation of momentum after the explosion. Since the

fragments have equal masses, the momentum in the x-direction is conserved.

Thus, we have:

m·v0x= 2m·v1x+ 2m·v2x

Where v1x= 8 m/s is the velocity of one fragment after the explosion and

v2xis the velocity of the other fragment.

Step 6: Solve for the velocity of the second fragment. Substitute known

values into the conservation of momentum equation and solve for v2x.

20 ·v0x= 2 ·8+2·v2x

20 ·20 ·cos(30◦) = 16 + 2 ·v2x

v2x= 140 m/s

Therefore, the speed of the other fragment just after the explosion is 140

m/s.

Question 12

Question

A baseball player hits a ball at an angle of 30◦above the horizontal with an

initial speed of 30 m/s. Calculate the maximum height the ball reaches and the

total time it is in the air.

Solution

Step 1: Break the initial velocity into its horizontal and vertical components.

The horizontal component (v0x) is given by:

v0x=v0cos(θ)

v0x= 30 ×cos(30◦)

v0x= 30 ×√3

v0x= 15√3 m/s

The vertical component (v0y) is given by:

v0y=v0sin(θ)

v0y= 30 ×sin(30◦)

v0y= 30 ×1

v0y= 15 m/s

Step 2: Calculate the time it takes to reach the highest point. The time

taken to reach the highest point is given by:

t=v0y

where gis the acceleration due to gravity (9.81 m/s2).

t=15

9.81

t≈1.53 s

Step 3: Calculate the maximum height reached. The maximum height

reached can be calculated using the equation:

y=v0yt−1

2gt2

Substitute v0y= 15 m/s, t= 1.53 s, and g= 9.81 m/s2.

y= 15 ×1.53 −1

2×9.81 ×(1.53)2

y≈22.9 m

Step 4: Calculate the total time the ball is in the air. The time of flight can

be calculated as twice the time taken to reach the highest point.

Total time = 2t

Total time = 2 ×1.53

Total time = 3.06 s

Therefore, the maximum height the ball reaches is approximately 22.9 m

and the total time it is in the air is 3.06 s.

Question 13

Question

A projectile is launched at an angle of 35◦above the horizontal with an initial

speed of 50 m/s. At the highest point in its trajectory, what are the projectile’s

horizontal and vertical velocities?

Solution

Step 1: Break the initial velocity into its horizontal and vertical components.

The initial velocity of the projectile can be broken down into horizontal and ver-

tical components: The horizontal component: vix=vicos(θ) = (50 m/s) cos(35◦)

The vertical component: viy=visin(θ) = (50 m/s) sin(35◦)

Step 2: Determine the vertical velocity at the highest point. Since at the

highest point the vertical velocity is momentarily zero, we have vfy= 0. The

vertical velocity at the highest point can be found using the equation: vfy=

viy−gt where gis the acceleration due to gravity (9.81 m/s2) and tis the time

taken to reach the highest point. Solve for tusing: 0 = (50 m/s) sin(35◦)−

(9.81 m/s2)t

Step 3: Calculate the time taken to reach the highest point. Solving the

above equation for t, we find: t=(50 m/s) sin(35◦)

9.81 m/s2

Step 4: Find the horizontal velocity at the highest point. The horizontal

velocity remains constant throughout the motion. Therefore, the horizontal

velocity at the highest point is the same as the initial horizontal velocity: vfx=

vix= (50 m/s) cos(35◦)

Step 5: Answer At the highest point in its trajectory, the projectile’s hori-

zontal velocity is (50 m/s) cos(35◦) and its vertical velocity is 0.

Question 14

Question

A soccer player kicks a ball at an angle of 30◦above the horizontal with an

initial speed of 20 m/s. The ball lands on the field 50 meters away. Find the

maximum height of the ball during its flight.

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components. The

initial velocity of the ball (v0) can be resolved into horizontal (v0x) and vertical

(v0y) components using trigonometry. v0x=v0cos(30◦) = 20 m/s ·cos(30◦)

v0y=v0sin(30◦) = 20 m/s ·sin(30◦)

Step 2: Determine the time of flight. The time of flight (t) can be found

from the vertical motion of the ball using the equation: y=v0yt−1

2gt2where

yis the vertical distance traveled (equal to the maximum height of the ball), g

is the acceleration due to gravity, and we can assume the ball lands at the same

height it was kicked from. Setting yto the maximum height and solving for t

gives us: 0 = v0yt−1

2gt2t=2v0y

Step 3: Calculate the maximum height. Substitute the expression for t

back into the equation for yto find the maximum height: ymax =v0y·2v0y

g−

2g2v0y

g2

Question 15

Question

A baseball is hit at an angle of 30◦above the horizontal with an initial speed

of 40 m/s. Calculate: (a) the time it takes for the baseball to reach the highest

point of its trajectory, and (b) the total horizontal distance traveled by the

baseball before hitting the ground.

Solution

(a) To find the time it takes for the baseball to reach the highest point, we can use

the fact that at the highest point the vertical component of velocity is zero. We

can use the equation for the vertical component of velocity: vy=voy −gt, where

vyis the vertical component of velocity, voy is the initial vertical component of

velocity, gis the acceleration due to gravity, and tis the time.

Step 1: Identify the known values. The initial vertical component of velocity

is given by voy =vosin θ= 40 sin 30◦= 20 m/s, and the acceleration due to

gravity is g= 9.81 m/s2.

Step 2: Find the time it takes to reach the highest point. Using vy=voy −gt

and vy= 0, we have: 0 = 20 −9.81t. Solve for t:t=20

9.81 ≈2.04 s.

Therefore, it takes approximately 2.04 seconds for the baseball to reach the

highest point of its trajectory.

(b) To find the total horizontal distance traveled by the baseball before

hitting the ground, we can use the equation for horizontal distance: x=voxt,

where xis the horizontal distance, vox is the initial horizontal component of

velocity, and tis the time taken.

Step 3: Find the horizontal component of velocity. The initial horizontal

component of velocity is given by vox =vocos θ= 40 cos 30◦= 34.64 m/s.

Step 4: Calculate the total horizontal distance. Using x=voxtand t= 2.04

s, we have: x= 34.64 ×2.04 ≈70.62 m.

Therefore, the total horizontal distance traveled by the baseball before hit-

ting the ground is approximately 70.62 meters.

Question 16

Question

A ball is kicked from the ground at an angle of 30◦above the horizontal with

an initial speed of 20 m/s. Calculate the maximum height reached by the ball.

Assume air resistance is negligible and g= 9.8 m/s2.

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components.

Step 2: Find the time taken to reach the maximum height. Step 3: Calculate

the maximum height reached by the ball. Step 4: Check your answer with an

alternative approach.

Step 1: The initial velocity v0can be resolved into horizontal and ver-

tical components as follows: Horizontal component: v0x=v0cos 30◦Vertical

component: v0y=v0sin 30◦

Given v0= 20 m/s, we can calculate v0xand v0y:v0x= 20 cos 30◦≈17.32

m/s v0y= 20 sin 30◦≈10 m/s

Step 2: The time taken for the ball to reach the maximum height can be

found using the vertical component of the initial velocity. The equation for the

vertical velocity as a function of time is: vy=v0y−gt

At the maximum height, the vertical velocity is zero. So, solving for t:

0 = 10 −9.8t t =10

9.8≈1.02 s

Step 3: The maximum height hmax can be calculated using the vertical

displacement formula: hmax =v0yt−1

2gt2

Substitute v0y,t, and ginto the equation: hmax = 10×1.02−1

2×9.8×(1.02)2

hmax ≈5.1 m

Therefore, the maximum height reached by the ball is approximately 5.1

meters.

Step 4: An alternative approach to solving for the maximum height is using

the formula: hmax =v0sin θ

g2

Substitute v0= 20 m/s, θ= 30◦, and g= 9.8 m/s2into the formula:

hmax =20 sin 30◦

9.82

hmax ≈5.1 m

This confirms that the maximum height reached by the ball is approximately

5.1 meters.

Question 17

Question

A projectile is launched from point A at an angle of 30◦above the horizontal

with an initial speed of 20 m/s. At the maximum height of its trajectory at

point B, the projectile explodes into two fragments. One fragment, fragment 1,

moves vertically upward with a speed of 10 m/s and reaches a maximum height

at point C. The other fragment, fragment 2, moves horizontally with a speed of

15 m/s and lands on the ground at point D. Calculate the distances AB, BC,

and CD.

Solution

Step 1: Calculate the time to reach the maximum height (point B) for the entire

projectile. We can use the kinematic equation for vertical motion:

y=vit−1

2gt2

At the maximum height, the vertical velocity is 0 m/s, so:

0 = 20 sin(30◦)t−1

2(9.8)t2

t=20 sin(30◦)

9.8

t≈1.02 s

Step 2: Calculate the height AB. Using the same kinematic equation:

y= 20 sin(30◦)t−1

2(9.8)t2

y= 20 sin(30◦)·1.02 −1

2(9.8)(1.02)2

y≈10.7 m

So, AB = 10.7 m.

Step 3: Calculate the time to reach the maximum height (point C) for frag-

ment 1. For fragment 1, the initial vertical velocity is 10 m/s, and acceleration

is -9.8 m/s2. Using the kinematic equation:

y=vit−1

2gt2

At the maximum height, the vertical velocity is 0 m/s:

0 = 10 −9.8t

t=10

9.8

t≈1.02 s

Step 4: Calculate the height BC. Using the same kinematic equation for

fragment 1:

y= 10t−1

2(9.8)t2

y= 10 ·1.02 −1

2(9.8)(1.02)2

y≈5.1 m

So, BC = 5.1 m.

Step 5: Calculate the time CD for fragment 2. Fragment 2 is moving hor-

izontally, so it will take the same time to reach point C as fragment 1 (1.02

s).

Step 6: Calculate the distance CD. The horizontal distance CD is the speed

of fragment 2 multiplied by the time:

CD = 15 ×1.02

CD ≈15.3 m

Therefore, the distances are: AB = 10.7 m, BC = 5.1 m, and CD = 15.3 m.

Question 18

Question

A baseball player throws a ball with an initial velocity of 20 m/s at an angle of

45◦above the horizontal.

1. Determine the time the ball is in the air.

2. Calculate the maximum height the ball reaches.

(Note: Use g= 9.81 m/s2for the acceleration due to gravity)

Solution

1. To determine the time the ball is in the air, we can analyze the vertical

motion of the ball. The vertical component of the initial velocity can be found

using viy =visin(θ). Here, vi= 20 m/s and θ= 45◦.

viy = 20 sin(45◦) = 20 √2

2!= 10√2 m/s

The ball is thrown upwards, so the acceleration is in the negative ydirection.

We can use the equation vf=vi+at to find the time the ball is in the air. At

its highest point, the ball’s final velocity will be 0.

0 = 10√2−9.81t

t=10√2

9.81 ≈1.44 s

2. To find the maximum height the ball reaches, we can use the kinematic

equation yf=yi+viyt−1

2gt2. At the highest point, the final vertical displace-

ment is 0.

0 = 0 + 10√2(1.44) −1

2(9.81)(1.44)2

0 = 14.4−9.81(2.0736)

0 = 14.4−20.35

20.35 = 14.4 + ymax

ymax = 20.35 −14.4=5.95 m

Therefore, the time the ball is in the air is approximately 1.44 s, and the

maximum height the ball reaches is 5.95 m.

Question 19

Question

A baseball is hit with an initial velocity of 30 m/s at an angle of 45◦above the

horizontal. Calculate the maximum height reached by the baseball. Assume air

resistance is negligible and take the acceleration due to gravity to be −9.8 m/s2.

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The initial velocity can be broken down into its horizontal component (vix) and

vertical component (viy). These components can be found using trigonometric

functions:

vix=vicos(θ)

viy=visin(θ)

where vi= 30 m/s is the initial velocity and θ= 45◦is the angle above the

horizontal. Substitute the given values:

vix= 30 m/s ·cos(45◦)

viy= 30 m/s ·sin(45◦)

Solving for vixand viy:

vix= 30 m/s ·√2

2= 15√2 m/s

viy= 30 m/s ·√2

2= 15√2 m/s

Step 2: Calculate the time taken to reach the maximum height. The time

taken for the baseball to reach the maximum height can be found using the

formula:

vf=vi+at

where vf= 0 m/s is the final vertical velocity at the maximum height, vi=

15√2 m/s is the initial vertical velocity, and a=−9.8 m/s2is the acceleration

due to gravity. Substitute the given values:

0 = 15√2−9.8t

Solving for t:

t=15√2

9.8≈2.14 s

Step 3: Calculate the maximum height reached by the baseball. The maxi-

mum height (h) can be found using the formula:

h=viyt+1

2at2

Substitute the known values:

h= 15√2·2.14 −1

2·9.8·(2.14)2

Solving for h:

h≈21.4 m

Therefore, the maximum height reached by the baseball is approximately

21.4 meters.

Question 20

Question

A projectile is fired from the ground at an angle of 30◦above the horizontal with

an initial speed of 40 m/s. At the highest point in its trajectory, the projectile

explodes into two fragments. One fragment, with mass 2 kg, lands 40 m from

the point directly below the explosion 10 seconds later. Determine the mass of

the other fragment and the horizontal and vertical components of the velocity

of both fragments at the time of the explosion.

Solution

Step 1: Find the time to reach the highest point using the y-component of the

velocity. Given: Initial speed v0= 40 m/s, Angle of projection θ= 30◦,g= 9.8

m/s2.

The ycomponent of the initial velocity is:

viy=v0sin θ.

Using the kinematic equation vf=vi+at and vf= 0, we can solve for the time

to reach the highest point:

0 = viy+ (−g)t.

Solving for tgives:

t=viy

g=40 ×sin 30◦

9.8=40 ×0.5

9.8= 2.04 s.

Step 2: Find the vertical component of the fragment’s velocity at the highest

point. At the highest point, the vertical component of the velocity is zero. So,

we can find the ycomponent of the velocity using vfy=viy−gt:

vfy= 40 ×0.5−9.8×2.04 = 10.2−19.98 = −9.78 m/s.

Step 3: Find the horizontal component of the fragment’s velocity at the time

of the explosion. The horizontal component of the velocity remains constant

throughout the trajectory. So, the horizontal component of the velocity at the

time of explosion is the same as the initial horizontal component:

vfx=vix=v0cos θ= 40 ×cos 30◦= 40 ×0.866 = 34.64 m/s.

Step 4: Use the information to find the mass of the other fragment. Let

the mass of the other fragment be m. The horizontal distance it travels before

landing is 40 m. We can use the equation of motion

x=vixt

to find the time it takes for the other fragment to land:

40 = 34.64 ×t.

Thus,

t=40

34.64 = 1.155 s.

Step 5: Find the vertical component of the other fragment’s velocity at the

time of the explosion. Using vfy=viy−gt:

vfy= 40 ×0.5−9.8×1.155 = 10.2−11.31 = −1.11 m/s.

Therefore, the mass of the other fragment is 2 kg, and at the time of ex-

plosion, the horizontal component of its velocity is 34.64 m/s and the vertical

component is -1.11 m/s.

Question 21

Question

A baseball player throws a ball from the outfield towards home plate. The ball

leaves the player’s hand at an angle of 30◦above the horizontal with an initial

speed of 30 m/s. Given that the distance between the pitcher’s mound and

home plate is 20 meters, determine whether the ball will reach home plate on

the fly.

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

With the angle of 30◦, the initial velocity components are:

v0x=v0cos θ= 30 cos 30◦≈26.0 m/s

v0y=v0sin θ= 30 sin 30◦≈15.0 m/s

Step 2: Determine the time taken for the ball to reach the home plate using

the vertical motion equation:

y=v0yt−1

2gt2

0 = 15t−1

2(9.81)t2

The solution to this quadratic equation is t≈3.06 s.

Step 3: Calculate the horizontal distance the ball travels using the horizontal

motion equation:

x=v0xt

x= 26.0×3.06 ≈79.6 m

Since the distance between the pitcher’s mound and home plate is 20 meters,

the ball will sail over home plate and continue its trajectory beyond the field.

Question 22

Question

A projectile is launched from the ground at an angle of 30 degrees above the

horizontal with an initial speed of 20 m/s. Determine the time it takes for the

projectile to reach its maximum height.

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components.

The initial velocity can be broken down into its horizontal and vertical compo-

nents: The initial horizontal velocity (vix) is given by:

vix=vi·cos(θ)

vix= 20 m/s ·cos(30◦)

vix= 20 m/s ·√3

vix= 10√3 m/s

The initial vertical velocity (viy) is given by:

viy=vi·sin(θ)

viy= 20 m/s ·sin(30◦)

viy= 20 m/s ·1

viy= 10 m/s

Step 2: Determine the time to reach maximum height.

The time it takes for the projectile to reach its maximum height can be calcu-

lated using the vertical component of the velocity and the acceleration due to

gravity. The equation to use is:

vf=viy−g·t

where vfis the final vertical velocity (0 m/s at maximum height), viyis the

initial vertical velocity (10 m/s), gis the acceleration due to gravity (-9.8 m/s2),

and tis the time.

Solving for time:

0 = 10 −9.8t

9.8t= 10

t=10

9.8

t≈1.02 s

Therefore, it takes approximately 1.02 seconds for the projectile to reach its

maximum height.

Question 23

Question

A projectile is launched from ground level with an initial speed of 30 m/s at

an angle of 60◦above the horizontal. At the highest point of its trajectory,

the projectile explodes into two pieces of equal mass. One piece comes straight

down with no initial speed. Where does the other piece land relative to the

point directly below the point of explosion?

Solution

Step 1: Solve for the time of flight and the maximum height reached by the

projectile.

Let’s consider the horizontal and vertical components of the projectile mo-

tion separately.

In the vertical direction: - The projectile starts and ends at the same height

(y= 0 at both points). - The initial vertical velocity is viy= 30 sin 60◦= 15√3

m/s. - The vertical acceleration is ay=−9.8 m/s2(due to gravity).

Using the kinematic equation vf=vi+a·t, where vf= 0 at the highest

point of the trajectory:

0 = 15√3−9.8·t

Solving for the time t, we get:

t=15√3

9.8≈2.73 s

The time of flight is twice this value:

Time of flight = 2 ·2.73 = 5.46 s

The maximum height ymax reached can be calculated using the kinematic

equation ymax =viy·t+1

2ayt2:

ymax = 15√3·2.73 + 1

2·(−9.8) ·(2.73)2≈37.5 m

Step 2: Determine the horizontal distance traveled by the projectile.

The horizontal velocity remains constant at vix= 30 cos 60◦= 15 m/s.

The horizontal distance xtraveled can be found using the equation x=vix·t:

x= 15 ·5.46 ≈81.9 m

Therefore, the projectile lands 81.9 meters away horizontally from the launch

point.

Step 3: Determine the landing point of the other piece.

Since one piece comes straight down with no initial speed, it will hit the

ground directly below the point of explosion. Hence, the other piece will also

land at a horizontal distance of 81.9 meters from the point of explosion.

Question 24

Question

A baseball player hits a ball at an angle of 30◦above the horizontal with an

initial velocity of 40 m/s. The ball lands on the flat roof of a 10 m high building

100 m away. What is the time of flight of the ball?

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components. The

initial velocity can be decomposed into its horizontal and vertical components:

v0x=v0cos θ

v0y=v0sin θ

Substitute v0= 40 m/s and θ= 30◦into the above equations to get:

v0x= 40 cos 30◦= 40 ·√3

2= 20√3 m/s

v0y= 40 sin 30◦= 40 ·1

2= 20 m/s

Step 2: Calculate the time taken for the ball to reach the roof. The vertical

motion of the ball can be analyzed using the following kinematic equation:

y=v0yt+1

2ayt2

We know that y= 10 m (height of the building), v0y= 20 m/s, ay=−9.8

m/s2(acceleration due to gravity in the vertical direction), and we want to find

t. Plugging these values into the equation gives:

10 = 20t−1

2·9.8·t2

10 = 20t−4.9t2

4.9t2−20t+ 10 = 0

Solving this quadratic equation will give us the time taken for the ball to

reach the roof.

Question 25

Question

A projectile is launched from the ground at an angle of 30◦above the horizontal

with an initial speed of 20 m/s. At a certain instant, the projectile is at a height

of 15 m above the ground. Find the magnitude of the velocity of the projectile

at this instant.

Solution

Step 1: Break the initial velocity into horizontal and vertical components. The

initial velocity of the projectile is given by v0= 20 m/s and the launch angle

is θ= 30◦. The horizontal component of the velocity is v0x=v0cos θand the

vertical component of the velocity is v0y=v0sin θ.

Step 2: Find the time taken for the projectile to reach a height of 15 m.

The vertical motion of the projectile can be described by the equation y=

y0+v0yt−1

2gt2, where yis the height of the projectile, y0is the initial height,

v0yis the initial vertical velocity, gis the acceleration due to gravity, and tis

the time. We know that y0= 0 m, y= 15 m, v0y= 20 sin 30◦= 10 m/s, and

g= 9.8 m/s2. Plugging these values in, we get:

15 = 10t−1

2(9.8)t2

Solving for t, we get t≈1.53 s.

Step 3: Find the velocity of the projectile at this instant. At the instant the

projectile is at a height of 15 m, we know the time t= 1.53 s. The horizontal

velocity remains constant, so vx=v0x= 20 cos 30◦= 17.32 m/s. The vertical

velocity at this instant can be found using the equation vy=v0y−gt. Plugging

in the values, we get:

vy= 10 −9.8(1.53) = −4.79 m/s

The magnitude of the velocity at this instant is:

v=qv2

x+v2

y=p17.322+ (−4.79)2≈17.89 m/s

Therefore, the magnitude of the velocity of the projectile at this instant is

approximately 17.89 m/s.

Question 26

Question

A projectile is launched with an initial speed of 40 m/s at an angle of 30 degrees

above the horizontal. Find the maximum height reached by the projectile.

Solution

Step 1: Break the initial velocity into its xand ycomponents. The initial speed

of 40 m/s can be broken into its xand ycomponents:

vix=vicos θ= 40 cos 30◦= 40 ·√3

2≈34.64 m/s

viy=visin θ= 40 sin 30◦= 40 ·1

2= 20 m/s

Step 2: Calculate the time to reach maximum height. The time to reach

maximum height can be found using the ycomponent of the initial velocity and

acceleration due to gravity.

vf=vi+at

At the peak of the projectile’s motion, the final velocity is 0 m/s.

0 = 20 −9.8t

t=20

9.8≈2.04 s

Step 3: Find the maximum height reached. The maximum height can be

calculated using the formula h=viyt−1

2gt2.

h= 20 ×2.04 −1

2×9.8×(2.04)2

h≈20.48 m

Therefore, the maximum height reached by the projectile is approximately

20.48 meters.

Question 27

Question

A projectile is launched from the ground at an angle of 45◦above the horizontal.

The initial velocity of the projectile is 30 m/s. Calculate the maximum height

reached by the projectile. (Neglect air resistance and take g= 9.81 m/s2)

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The initial velocity v0can be resolved into horizontal component v0xand vertical

component v0yas follows:

v0x=v0cos θ= 30 cos 45◦

v0y=v0sin θ= 30 sin 45◦

Step 2: Determine the time taken to reach the maximum height. The time

taken to reach the maximum height can be determined using the formula:

vfy =v0y−gt

At the maximum height, the vertical component of the velocity will be 0. So,

we have:

0 = v0y−gtmax

Solving for tmax gives:

tmax =v0y

Step 3: Calculate the maximum height. The maximum height hmax can be

calculated using the formula:

hmax =v0ytmax −1

2gt2

max

Substitute the values of v0yand tmax into the equation to find hmax.

Question 28

Question

A projectile is launched at an angle of 30◦above the horizontal with an initial

speed of 20 m/s. Calculate the maximum height reached by the projectile.

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components.

The horizontal component is given by: vix=vi·cos(θ) The vertical component

is given by: viy=vi·sin(θ), where vi= 20 m/s and θ= 30◦. So, vix=

20 m/s ·cos(30◦)≈17.32 m/s, viy= 20 m/s ·sin(30◦)≈10 m/s.

Step 2: Calculate the time taken to reach the maximum height. Use the

equation: vfy=viy−gt, where vfy= 0 m/s (at the maximum height) and

g= 9.81 m/s2. So, 0 = 10 −9.81t, which gives t≈1.02 s.

Step 3: Calculate the maximum height reached. Use the equation: ymax =

yi+viyt−1

2gt2, where yi= 0 (initial vertical position) and t= 1.02 s. So,

ymax = 0 + 10 ·1.02 −1

2·9.81 ·(1.02)2≈5.10 m.

Therefore, the maximum height reached by the projectile is approximately

5.10 meters.

Question 29

Question

A projectile is launched from the ground with an initial speed of 30 m/s at an

angle of 30◦above the horizontal. At what time after being launched does the

projectile reach a height of 10 meters?

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components. The

horizontal component of the initial velocity is given by v0x=v0cos θ, where v0

is the initial speed (30 m/s) and θis the angle of launch (30 degrees). Therefore,

v0x= 30 cos 30◦≈25.98 m/s

The vertical component of the initial velocity is given by v0y=v0sin θ, so

v0y= 30 sin 30◦= 15 m/s

Step 2: Determine the time it takes for the projectile to reach a height of 10

meters. Consider the motion of the projectile in the vertical direction. We will

use the kinematic equation:

y=y0+v0yt−1

2gt2

where y0is the initial height (0 meters), yis the final height (10 meters), v0yis

the initial vertical velocity (15 m/s), gis the acceleration due to gravity (-9.8

m/s2), and tis the time we need to find.

Plugging in the values we have:

10 = 0 + 15t−1

2(9.8)t2

Step 3: Solve for the time t. Rearranging the equation, we get:

4.9t2−15t+ 10 = 0

This is a quadratic equation that can be solved using the quadratic formula.

The solutions are:

t=15 ±p152−4(4.9)(10)

2(4.9)

t≈3.35 s or 0.60 s

Since the time cannot be negative, the projectile reaches a height of 10 meters

at approximately 0.60 seconds after being launched.

Question 30

Question

A ball is kicked with an initial velocity of 20 m/s at an angle of 30◦above the

horizontal. Find the maximum height reached by the ball.

Solution

Step 1: Break the initial velocity into its horizontal and vertical components.

The horizontal component, v0x, is given by v0x=v0cos(θ), where v0=

20 m/s and θ= 30◦. Therefore,

v0x= 20 m/s ×cos(30◦) = 20 m/s ×√3

2= 10√3 m/s

The vertical component, v0y, is given by v0y=v0sin(θ), where v0= 20 m/s

and θ= 30◦. Therefore,

v0y= 20 m/s ×sin(30◦) = 20 m/s ×1

2= 10 m/s

Step 2: Use the vertical component to find the time taken to reach the

maximum height.

The ball reaches its maximum height when the vertical velocity becomes 0.

We can use the equation vf=v0y−gt where vf= 0 and g= 9.8 m/s2.

0 = 10 −9.8t

t=10

9.8≈1.02 s

Step 3: Calculate the maximum height using the vertical component and

time.

The maximum height Hreached by the ball can be found using the equation

H=v0yt−1

2gt2.

H= 10 ×1.02 −1

2×9.8×(1.02)2

H≈5.1−5≈0.1 m

Therefore, the maximum height reached by the ball is approximately 0.1 m.

Question 31

Question

A projectile is launched from the ground at an angle of 30◦above the horizontal

with an initial speed of 20 m/s. At the highest point of the trajectory, the

projectile explodes into two fragments. One fragment, of mass 2 kg, moves

vertically upward with a speed of 4 m/s immediately after the explosion, while

the other fragment, of mass 3 kg, moves horizontally with a speed of 10 m/s.

Calculate the height above the ground at which the explosion occurs.

Solution

Step 1: Calculate the time of flight of the projectile motion.

The time of flight is given by:

T=2Visin θ

where Viis the initial velocity, θis the launch angle, and gis the acceleration

due to gravity.

Given Vi= 20 m/s, θ= 30◦, and g= 9.81 m/s2, we can calculate T:

T=2×20 ×sin(30◦)

9.81 ≈2.04 s

Step 2: Calculate the height at the top of the trajectory.

The maximum height Hreached by the projectile is given by:

H=V2

isin2θ

Plugging in the known values:

H=(20)2×(sin 30◦)2

2×9.81 ≈5.10 m

Step 3: Set up and solve simultaneous equations for the vertical and hori-

zontal components of the explosion.

Let hbe the height at which the explosion occurs. For the vertical motion,

we have:

0−h= 4t−1

2gt2

where tis the time taken for the projectile to reach height h.

For the horizontal motion, we have:

0−0 = 10t

The times for both the vertical and horizontal motion are the same, so we can

solve for tfrom the horizontal motion equation:

t=h

Substitute this time into the vertical motion equation:

0−h= 4 ×h

10 −1

2gh

102

Solving this equation gives the height of the explosion, h:

h= 15.6 m

Therefore, the height above the ground at which the explosion occurs is 15.6

Question 32

Question

A particle moves along a curved path in the xy plane according to the following

equations: x(t) = 3t2−2tand y(t) = 4t3−t. Find the particle’s velocity and

acceleration vectors at t= 2 seconds.

Solution

Step 1: To find the particle’s velocity, we first calculate the derivatives of the

position functions x(t) and y(t) to obtain the x(t) and y(t) components of ve-

locity.

Given x(t)=3t2−2tand y(t) = 4t3−t, the velocity components are:

vx(t) = dx

dt =d

dt (3t2−2t)=6t−2

vy(t) = dy

dt =d

dt (4t3−t) = 12t2−1

At t= 2 s, the velocity vector is:

v=vx(2)i+vy(2)j= 10i+ 47jm/s

Step 2: Next, we find the particle’s acceleration vector by taking the deriva-

tives of the velocity components.

The acceleration components are:

ax(t) = dvx

dt =d

dt (6t−2) = 6

ay(t) = dvy

dt =d

dt (12t2−1) = 24t

At t= 2 s, the acceleration vector is:

a=ax(2)i+ay(2)j= 6i+ 48jm/s2

Question 33

Question

A particle starts at the origin of an xy-coordinate system and moves in the

xy-plane with velocity components vx= 3tm/s and vy= 4 −2tm/s, where tis

in seconds. Determine the magnitude and direction of the particle’s velocity at

t= 2 s.

Solution

Step 1: Recall that the velocity of a particle in two dimensions is given by

v =vxˆ

i+vyˆ

j, where vxis the velocity component in the x-direction, and vyis

the velocity component in the y-direction.

Given vx= 3tm/s and vy= 4 −2tm/s, we have v = 3tˆ

i+ (4 −2t)ˆ

Step 2: To find the magnitude of the velocity, we use the formula |v|=

qv2

x+v2

Substitute the given values at t= 2 s:

|v|=p(3(2))2+ (4 −2(2))2

|v|=√36 + 0

|v|=√36

|v|= 6 m/s

Step 3: To find the direction of the velocity, we use the formula θ=

tan−1vy

vx.

Substitute the given values at t= 2 s:

θ= tan−14−2(2)

3(2) 

θ= tan−10

6

θ= tan−1(0)

θ= 0◦

Therefore, at t= 2 s, the magnitude of the particle’s velocity is 6 m/s, and

its direction is in the positive x-direction.

Question 34

Question

A baseball is hit into the outfield at an angle of 35◦above the horizontal with

an initial speed of 30 m/s. How far from home plate does the baseball land if it

is caught by an outfielder 1.5 m above the ground?

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The initial velocity of the baseball can be resolved into its horizontal and vertical

components as follows:

V0x=V0cos θ= 30 cos 35◦≈24.69 m/s

V0y=V0sin θ= 30 sin 35◦≈17.21 m/s

Step 2: Calculate the time it takes for the baseball to reach the outfielder.

The time taken for the baseball to reach the outfielder can be calculated using

the vertical component of the motion with the equation:

y=V0yt+1

2gt2

where y= 1.5 m (height of the outfielder) and g= 9.81 m/s2(acceleration due

to gravity). Solving for t:

1.5 = 17.21t−4.905t2

4.905t2−17.21t+ 1.5=0

Using the quadratic formula, we find the positive value for tto be approximately

t≈1.42 s.

Step 3: Calculate the horizontal distance the baseball travels before reaching

the outfielder. The horizontal distance can be calculated using the equation:

x=V0xt

x= 24.69 ×1.42 ≈35.06 m

Therefore, the baseball lands approximately 35.06 meters away from home

plate.

Question 35

Question

A baseball is hit at an angle of 35◦above the horizontal with an initial speed

of 30 m/s from a height of 2 m above the ground. Ignoring air resistance,

determine the maximum height the baseball reaches and the total time it’s in

the air.

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The initial velocity of the baseball can be resolved into horizontal and vertical

components as follows:

vix=vicos(θ) = 30 cos(35◦)≈24.63 m/s

viy=visin(θ) = 30 sin(35◦)≈17.14 m/s

Step 2: Determine the time it takes to reach the maximum height.

To find the time it takes to reach the maximum height, we use the vertical

equation of motion:

y=yi+viyt−1

2gt2

where yi= 2 m is the initial height, yis the maximum height (which occurs

when the vertical velocity becomes 0), and g= 9.81 m/s2is the acceleration

due to gravity.

At the maximum height, vy= 0, so:

0 = 17.14 −9.81t

Using the kinematic equation: vf=vi,y +aytwhere: vf= 0 (final vertical

velocity) vi,y = 10 m/s (initial vertical velocity) ay=−9.81 m/s2(acceleration

due to gravity) t= 2.04 s (time of flight)

Substitute the given values: 0 = 10−9.81 ×2.04 0 = 10 −19.9724 19.9724 =

The stone cannot reach such a height. Let’s recalculate with the correct

formula.

The height hof the cliff can be calculated using the horizontal motion of the

stone and the time of flight. The horizontal distance traveled by the stone is:

x=vi,xt

Given: vi,x = 10√3 m/s t= 4 s

Substitute these values: x= 10√3×4 = 40√3 m

Since the stone falls from the top of the cliff to the ground, the height of

the cliff is equal to the vertical position at t= 0. Using the formula: y=

vi,yt+1

2ayt2where: y=h(height of the cliff) vi,y = 10 m/s (initial vertical

velocity) ay=−9.81 m/s2(acceleration due to gravity) t= 4 s

Substitute the given values and solve for

Question 2

Question

A projectile is launched from the ground with an initial speed of 50 m/s at an

angle of 30◦above the horizontal. Find the total time the projectile is in the

air before hitting the ground.

Solution

Step 1: Resolve the initial velocity vector into its horizontal and vertical compo-

nents. The initial velocity vector v0can be expressed as the sum of its horizontal

component v0xand its vertical component v0y:

v0x=v0cos θ= 50 ×cos(30◦) = 43.3 m/s

v0y=v0sin θ= 50 ×sin(30◦) = 25 m/s

Step 2: Calculate the time of flight. The time of flight can be found using

the vertical motion equation:

y=v0yt+1

2ayt2

where yis the vertical displacement (final height - initial height) and ayis the

acceleration due to gravity (-9.81 m/s2). Substitute y= 0 (since the projectile

returns to the initial height) and solve for t:

0 = 25t−4.91t2

4.91t2−25t= 0

t(4.91t−25) = 0

t= 0 s or t=25

4.91 ≈5.09 s

Step 3: Double the time of flight to find the total time in the air. Since the

projectile is in the air for the same duration both upward and downward, the

total time in the air is:

Total time = 2 ×5.09 s = 10.18 s

Therefore, the total time the projectile is in the air before hitting the ground

is approximately 10.18 seconds.

Question 3

Question

A cannonball is fired at an angle of 30◦above the horizontal from the top of

a cliff that is 80 meters high. If the initial speed of the cannonball is 40 m/s,

determine the time it takes for the cannonball to hit the ground. Ignore air

resistance and assume g= 9.8 m/s2.

Solution

Step 1: Resolve the initial velocity of the cannonball into horizontal and vertical

components. Let v0= 40 m/s be the initial speed, θ= 30◦be the angle above the

horizontal, v0xbe the horizontal component, and v0ybe the vertical component.

v0x=v0cos(θ)

= 40 m/s ·cos(30◦)

= 40 m/s ·√3

= 20√3 m/s

v0y=v0sin(θ) = 40 m/s ·sin(30◦) = 40 m/s ·1

2= 20 m/s

Step 2: Determine the time it takes for the cannonball to reach the ground

using the vertical component. The vertical equation of motion is given by:

y=v0yt+1

2gt2

Since the cannonball starts at a height of 80 meters above the ground, when

it hits the ground, y=−80 meters. Thus, the vertical equation becomes:

−80 = 20t−1

2·9.8·t2

−80 = 20t−4.9t2

Solving for t:

4.9t2−20t−80 = 0

Using the quadratic formula, we find:

t=−(−20) ±p(−20)2−4·4.9·(−80)

2·4.9

t=20 ±√400 + 1568

9.8

t=20 ±√1968

9.8

t≈20 ±44.36

9.8

which gives us:

t1≈20 + 44.36

9.8=64.36

9.8≈6.57 s

t2≈20 −44.36

9.8=−24.36

9.8=−2.49 s

Since time cannot be negative, the time it takes for the cannonball to hit

the ground is approximately 6.57 seconds.

Question 4

Question

A particle is projected from the origin with an initial velocity of 30 m/s at an

angle of 30◦above the horizontal. Calculate the time when the particle is at

a height of 15 m above the origin and determine the horizontal distance it has

traveled at that time.

Solution

Let’s first analyze the motion of the particle in the vertical direction and then

in the horizontal direction.

Vertical Direction:



Given: v0y= 30 m/s ·sin(30◦) = 15 m/s, ay=−9.8 m/s2, and yf= 15 m.



We will use the kinematic equation yf=y0+v0yt+1

2ayt2to solve for the

time t.



Substituting the given values, we have 15 = 0 + 15t−1

2·9.8t2.



This equation simplifies to 4.9t2−15t+ 15 = 0.



We can solve this quadratic equation to find the positive value of t.

Horizontal Direction:



Given: v0x= 30 m/s ·cos(30◦) = 26 m/s.



The horizontal distance xtraveled by the particle at time tis given by

x=v0xt.



We will substitute v0xand the calculated value of tto find x.

Question 5

Question

A baseball player throws a ball with an initial velocity of 20 m/s at an angle of

30 degrees above the horizontal. The ball lands on the ground 100 meters away

from the player. Find the maximum height reached by the ball during its flight.

Solution

Step 1: Resolve the initial velocity of the ball into horizontal and vertical com-

ponents. The horizontal component of the initial velocity (v0x) can be found

using:

v0x=v0cos(θ)

where v0= 20 m/s and θ= 30◦.

v0x= 20 m/s ×cos(30◦)

v0x= 20 m/s ×√3

v0x= 10√3 m/s

The vertical component of the initial velocity (v0y) can be found using:

v0y=v0sin(θ)

v0y= 20 m/s ×sin(30◦)

v0y= 20 m/s ×1

v0y= 10 m/s

Step 2: Find the time taken to reach the peak height. Since the vertical

motion of the ball is symmetrical, the time taken to reach the peak height is

half of the total time of flight. The equation for the vertical motion of the ball

is:

y=v0yt−1

2gt2

where yis the maximum height reached. At the peak height, the vertical velocity

becomes 0.

0 = v0y−gt

t=v0y

t=10 m/s

9.81 m/s2

t≈1.02 s

Step 3: Find the maximum height reached by the ball. Substitute the value

of tinto the equation for vertical motion:

y= 10 m/s ×1.02 s −1

2×9.81 m/s2×(1.02 s)2

y= 10.2 m −1

2×9.81 m/s2×1.04 s2

y= 10.2 m −1

2×9.81 m/s2×1.04 s2

y= 10.2 m −5.088 m

y≈5.112 m

Therefore, the maximum height reached by the ball during its flight is ap-

proximately 5.112 meters.

Question 6

Question

A baseball is hit with an initial velocity of 20 m/s at an angle of 30◦above the

horizontal. Ignoring air resistance, calculate the time it takes for the baseball

to reach its maximum height.

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The horizontal component is given by:

v0x=v0cos(θ) = 20 cos(30◦)≈17.32 m/s

The vertical component is given by:

v0y=v0sin(θ) = 20 sin(30◦)≈10 m/s

Step 2: Determine the time it takes to reach maximum height.

The baseball reaches maximum height when the vertical component of its

velocity is 0 m/s. We can use the equation vy=v0y−gt, where vyis the vertical

component of velocity at a given time t,v0yis the initial vertical component of

velocity, gis the acceleration due to gravity (9.8 m/s2), and tis the time taken.

Setting vy= 0 and v0y= 10 m/s, we get:

0 = 10 −9.8t

t=10

9.8≈1.02 s

Therefore, it takes approximately 1.02 seconds for the baseball to reach its

maximum height.

Question 7

Question

A football quarterback throws a pass to a stationary receiver who is 20 meters

away. The initial velocity of the football is 15 m/s at an angle of 30◦above the

horizontal. What is the maximum height above the quarterback’s hand that

the ball reaches during its flight?

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

Let v0= 15 m/s be the initial velocity and θ= 30◦be the angle above the hor-

izontal. The horizontal component is given by v0x=v0cos(θ) and the vertical

component is given by v0y=v0sin(θ). Therefore, v0x= 15 cos(30◦) = 13.0 m/s

and v0y= 15 sin(30◦)=7.5 m/s.

Step 2: Determine the time of flight. Using the vertical component of ve-

locity, we can find the time it takes for the ball to reach its maximum height

using the equation vf=vi+at, where a=−9.8 m/s2is the acceleration due to

gravity and vf= 0 m/s at the maximum height. So, 0 = 7.5−9.8twhich gives

t= 0.77 s.

Step 3: Calculate the maximum height. The maximum height hcan be

found using the equation h=v0yt+1

2at2. Substitute v0y= 7.5 m/s, t= 0.77 s,

and a=−9.8 m/s2to get: h= 7.5(0.77) + 1

2(−9.8)(0.77)2= 5.77 m.

Therefore, the maximum height above the quarterback’s hand that the ball

reaches during its flight is 5.77 meters.

Question 8

Question

A baseball is hit at an angle of 30◦above the horizontal with an initial speed of

25 m/s. What are the horizontal and vertical components of its velocity at the

highest point of its trajectory?

Solution

Step 1: Identify the given information.

The initial speed of the baseball is v0= 25 m/s and the launch angle is θ= 30◦.

We will need to find the horizontal and vertical components of the velocity at

the highest point of its trajectory.

Step 2: Break the initial velocity into horizontal and vertical components.

The initial velocity can be broken down into its horizontal and vertical compo-

nents using trigonometric functions. The horizontal component can be found

using v0x=v0cos(θ) where θ= 30◦:

v0x= 25 m/s ·cos(30◦)

v0x= 25 ·√3

2≈21.65 m/s

The vertical component can be found using v0y=v0sin(θ) where θ= 30◦:

v0y= 25 m/s ·sin(30◦)

v0y= 25 ·1

2= 12.5 m/s

Step 3: Determine the horizontal and vertical components of velocity at the

highest point.

At the highest point of the trajectory, the vertical component of the velocity

is momentarily zero. Therefore, the horizontal component remains constant at

v0xand the vertical component becomes 0 at the highest point.

Thus, at the highest point of the trajectory: Horizontal component (vhx)

= 21.65 m/s Vertical component (vhy) = 0

Question 9

Question

A projectile is launched with an initial speed of 30 m/s at an angle of 60◦above

the horizontal. Find the time to reach maximum height, the maximum height

above the launch point, and the total time in the air.

Solution

Step 1: Break the initial velocity into its horizontal and vertical components.

The initial velocity of the projectile can be broken down into its horizontal and

vertical components as follows:

vix=vicos θ= 30 cos 60◦= 15 m/s

viy=visin θ= 30 sin 60◦= 25√3 m/s

Step 2: Find the time to reach maximum height. The time to reach maxi-

mum height can be found using the equation vf=vi+at, where vf= 0 at the

maximum height and a=−9.8 m/s2is the acceleration due to gravity. Solving

for tyields:

0 = 25√3−9.8t

t=25√3

9.8≈5.03 s

Step 3: Find the maximum height above the launch point. The maximum

height above the launch point can be found using the kinematic equation ymax =

yi+viyt+1

2at2, where yi= 0 and a=−9.8 m/s2. Thus, we have:

ymax = 25√3×5.03 −0.5×9.8×(5.03)2≈63.45 m

Step 4: Find the total time in the air. The total time in the air can be found

by doubling the time it took to reach the maximum height. Thus,

Total time = 2 ×5.03 = 10.06 s

Therefore, the projectile will reach its maximum height after approximately

5.03 seconds, at a height of 63.45 meters, and will be in the air for a total of

approximately 10.06 seconds.

Question 10

Question

A projectile is launched from the ground with an initial velocity of 20 m/s at

an angle of 30◦above the horizontal. Find the time it takes for the projectile

to reach its maximum height.

Solution

Step 1: Break the initial velocity into its horizontal and vertical components.

The horizontal component of the initial velocity is given by:

V0x=V0cos(θ)

where V0xis the horizontal component of the initial velocity, V0is the initial

velocity, and θis the launch angle. Substitute V0= 20 m/s and θ= 30◦:

V0x= 20 m/s ·cos(30◦)

V0x= 20 m/s ·√3

V0x= 10√3 m/s

The vertical component of the initial velocity is given by:

V0y=V0sin(θ)

where V0yis the vertical component of the initial velocity. Substitute V0=

20 m/s and θ= 30◦:

V0y= 20 m/s ·sin(30◦)

V0y= 20 m/s ·1

V0y= 10 m/s

Step 2: Determine the time to reach maximum height. At the maximum

height, the vertical component of the velocity becomes zero. Using the kinematic

equation for vertical motion:

Vy=V0y−gt

where Vyis the vertical component of the velocity, gis the acceleration due to

gravity (9.8 m/s2), and tis the time. Substitute Vy= 0, V0y= 10 m/s, and

g= 9.8 m/s2:

0 = 10 −9.8t

t=10

9.8

t≈1.02 s

Therefore, the time it takes for the projectile to reach its maximum height

is approximately 1.02 seconds.

Question 11

Question

A projectile is launched from the ground at an angle of 30◦above the horizontal

with an initial speed of 20 m/s. At its highest point, the projectile explodes into

two fragments of equal mass. One fragment continues in the original direction

with a speed of 8 m/s. What is the speed of the other fragment just after the

explosion?

Solution

Step 1: Analyze the projectile motion before the explosion. Let’s denote the

initial speed of the projectile as v0= 20 m/s. The horizontal component of

the initial velocity is v0x=v0cos(30◦) and the vertical component is v0y=

v0sin(30◦).

Step 2: Find the time to reach the highest point. The time it takes for the

projectile to reach its highest point can be found using the vertical component

of the motion. The vertical component of the velocity at the highest point is

zero. We can use the equation vy=v0y−gt where vy= 0 to find the time t.

0 = v0y−gt =⇒t=v0y

Step 3: Calculate the maximum height. The maximum height Hreached by

the projectile can be found using the equation H=v0y·t−1

2gt2.

H=v0y·v0y

g−1

2gv0y

g2

Step 4: Determine the horizontal distance traveled. The horizontal distance

Rtraveled by the projectile before the explosion can be found using the equation

R=vx·(2t) since the projectile will reach the same height on its way down as

it did on its way up.

R=v0x·(2t)

Step 5: Apply the conservation of momentum after the explosion. Since the

fragments have equal masses, the momentum in the x-direction is conserved.

Thus, we have:

m·v0x= 2m·v1x+ 2m·v2x

Where v1x= 8 m/s is the velocity of one fragment after the explosion and

v2xis the velocity of the other fragment.

Step 6: Solve for the velocity of the second fragment. Substitute known

values into the conservation of momentum equation and solve for v2x.

20 ·v0x= 2 ·8+2·v2x

20 ·20 ·cos(30◦) = 16 + 2 ·v2x

v2x= 140 m/s

Therefore, the speed of the other fragment just after the explosion is 140

m/s.

Question 12

Question

A baseball player hits a ball at an angle of 30◦above the horizontal with an

initial speed of 30 m/s. Calculate the maximum height the ball reaches and the

total time it is in the air.

Solution

Step 1: Break the initial velocity into its horizontal and vertical components.

The horizontal component (v0x) is given by:

v0x=v0cos(θ)

v0x= 30 ×cos(30◦)

v0x= 30 ×√3

v0x= 15√3 m/s

The vertical component (v0y) is given by:

v0y=v0sin(θ)

v0y= 30 ×sin(30◦)

v0y= 30 ×1

v0y= 15 m/s

Step 2: Calculate the time it takes to reach the highest point. The time

taken to reach the highest point is given by:

t=v0y

where gis the acceleration due to gravity (9.81 m/s2).

t=15

9.81

t≈1.53 s

Step 3: Calculate the maximum height reached. The maximum height

reached can be calculated using the equation:

y=v0yt−1

2gt2

Substitute v0y= 15 m/s, t= 1.53 s, and g= 9.81 m/s2.

y= 15 ×1.53 −1

2×9.81 ×(1.53)2

y≈22.9 m

Step 4: Calculate the total time the ball is in the air. The time of flight can

be calculated as twice the time taken to reach the highest point.

Total time = 2t

Total time = 2 ×1.53

Total time = 3.06 s

Therefore, the maximum height the ball reaches is approximately 22.9 m

and the total time it is in the air is 3.06 s.

Question 13

Question

A projectile is launched at an angle of 35◦above the horizontal with an initial

speed of 50 m/s. At the highest point in its trajectory, what are the projectile’s

horizontal and vertical velocities?

Solution

Step 1: Break the initial velocity into its horizontal and vertical components.

The initial velocity of the projectile can be broken down into horizontal and ver-

tical components: The horizontal component: vix=vicos(θ) = (50 m/s) cos(35◦)

The vertical component: viy=visin(θ) = (50 m/s) sin(35◦)

Step 2: Determine the vertical velocity at the highest point. Since at the

highest point the vertical velocity is momentarily zero, we have vfy= 0. The

vertical velocity at the highest point can be found using the equation: vfy=

viy−gt where gis the acceleration due to gravity (9.81 m/s2) and tis the time

taken to reach the highest point. Solve for tusing: 0 = (50 m/s) sin(35◦)−

(9.81 m/s2)t

Step 3: Calculate the time taken to reach the highest point. Solving the

above equation for t, we find: t=(50 m/s) sin(35◦)

9.81 m/s2

Step 4: Find the horizontal velocity at the highest point. The horizontal

velocity remains constant throughout the motion. Therefore, the horizontal

velocity at the highest point is the same as the initial horizontal velocity: vfx=

vix= (50 m/s) cos(35◦)

Step 5: Answer At the highest point in its trajectory, the projectile’s hori-

zontal velocity is (50 m/s) cos(35◦) and its vertical velocity is 0.

Question 14

Question

A soccer player kicks a ball at an angle of 30◦above the horizontal with an

initial speed of 20 m/s. The ball lands on the field 50 meters away. Find the

maximum height of the ball during its flight.

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components. The

initial velocity of the ball (v0) can be resolved into horizontal (v0x) and vertical

(v0y) components using trigonometry. v0x=v0cos(30◦) = 20 m/s ·cos(30◦)

v0y=v0sin(30◦) = 20 m/s ·sin(30◦)

Step 2: Determine the time of flight. The time of flight (t) can be found

from the vertical motion of the ball using the equation: y=v0yt−1

2gt2where

yis the vertical distance traveled (equal to the maximum height of the ball), g

is the acceleration due to gravity, and we can assume the ball lands at the same

height it was kicked from. Setting yto the maximum height and solving for t

gives us: 0 = v0yt−1

2gt2t=2v0y

Step 3: Calculate the maximum height. Substitute the expression for t

back into the equation for yto find the maximum height: ymax =v0y·2v0y

g−

2g2v0y

g2

Question 15

Question

A baseball is hit at an angle of 30◦above the horizontal with an initial speed

of 40 m/s. Calculate: (a) the time it takes for the baseball to reach the highest

point of its trajectory, and (b) the total horizontal distance traveled by the

baseball before hitting the ground.

Solution

(a) To find the time it takes for the baseball to reach the highest point, we can use

the fact that at the highest point the vertical component of velocity is zero. We

can use the equation for the vertical component of velocity: vy=voy −gt, where

vyis the vertical component of velocity, voy is the initial vertical component of

velocity, gis the acceleration due to gravity, and tis the time.

Step 1: Identify the known values. The initial vertical component of velocity

is given by voy =vosin θ= 40 sin 30◦= 20 m/s, and the acceleration due to

gravity is g= 9.81 m/s2.

Step 2: Find the time it takes to reach the highest point. Using vy=voy −gt

and vy= 0, we have: 0 = 20 −9.81t. Solve for t:t=20

9.81 ≈2.04 s.

Therefore, it takes approximately 2.04 seconds for the baseball to reach the

highest point of its trajectory.

(b) To find the total horizontal distance traveled by the baseball before

hitting the ground, we can use the equation for horizontal distance: x=voxt,

where xis the horizontal distance, vox is the initial horizontal component of

velocity, and tis the time taken.

Step 3: Find the horizontal component of velocity. The initial horizontal

component of velocity is given by vox =vocos θ= 40 cos 30◦= 34.64 m/s.

Step 4: Calculate the total horizontal distance. Using x=voxtand t= 2.04

s, we have: x= 34.64 ×2.04 ≈70.62 m.

Therefore, the total horizontal distance traveled by the baseball before hit-

ting the ground is approximately 70.62 meters.

Question 16

Question

A ball is kicked from the ground at an angle of 30◦above the horizontal with

an initial speed of 20 m/s. Calculate the maximum height reached by the ball.

Assume air resistance is negligible and g= 9.8 m/s2.

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components.

Step 2: Find the time taken to reach the maximum height. Step 3: Calculate

the maximum height reached by the ball. Step 4: Check your answer with an

alternative approach.

Step 1: The initial velocity v0can be resolved into horizontal and ver-

tical components as follows: Horizontal component: v0x=v0cos 30◦Vertical

component: v0y=v0sin 30◦

Given v0= 20 m/s, we can calculate v0xand v0y:v0x= 20 cos 30◦≈17.32

m/s v0y= 20 sin 30◦≈10 m/s

Step 2: The time taken for the ball to reach the maximum height can be

found using the vertical component of the initial velocity. The equation for the

vertical velocity as a function of time is: vy=v0y−gt

At the maximum height, the vertical velocity is zero. So, solving for t:

0 = 10 −9.8t t =10

9.8≈1.02 s

Step 3: The maximum height hmax can be calculated using the vertical

displacement formula: hmax =v0yt−1

2gt2

Substitute v0y,t, and ginto the equation: hmax = 10×1.02−1

2×9.8×(1.02)2

hmax ≈5.1 m

Therefore, the maximum height reached by the ball is approximately 5.1

meters.

Step 4: An alternative approach to solving for the maximum height is using

the formula: hmax =v0sin θ

g2

Substitute v0= 20 m/s, θ= 30◦, and g= 9.8 m/s2into the formula:

hmax =20 sin 30◦

9.82

hmax ≈5.1 m

This confirms that the maximum height reached by the ball is approximately

5.1 meters.

Question 17

Question

A projectile is launched from point A at an angle of 30◦above the horizontal

with an initial speed of 20 m/s. At the maximum height of its trajectory at

point B, the projectile explodes into two fragments. One fragment, fragment 1,

moves vertically upward with a speed of 10 m/s and reaches a maximum height

at point C. The other fragment, fragment 2, moves horizontally with a speed of

15 m/s and lands on the ground at point D. Calculate the distances AB, BC,

and CD.

Solution

Step 1: Calculate the time to reach the maximum height (point B) for the entire

projectile. We can use the kinematic equation for vertical motion:

y=vit−1

2gt2

At the maximum height, the vertical velocity is 0 m/s, so:

0 = 20 sin(30◦)t−1

2(9.8)t2

t=20 sin(30◦)

9.8

t≈1.02 s

Step 2: Calculate the height AB. Using the same kinematic equation:

y= 20 sin(30◦)t−1

2(9.8)t2

y= 20 sin(30◦)·1.02 −1

2(9.8)(1.02)2

y≈10.7 m

So, AB = 10.7 m.

Step 3: Calculate the time to reach the maximum height (point C) for frag-

ment 1. For fragment 1, the initial vertical velocity is 10 m/s, and acceleration

is -9.8 m/s2. Using the kinematic equation:

y=vit−1

2gt2

At the maximum height, the vertical velocity is 0 m/s:

0 = 10 −9.8t

t=10

9.8

t≈1.02 s

Step 4: Calculate the height BC. Using the same kinematic equation for

fragment 1:

y= 10t−1

2(9.8)t2

y= 10 ·1.02 −1

2(9.8)(1.02)2

y≈5.1 m

So, BC = 5.1 m.

Step 5: Calculate the time CD for fragment 2. Fragment 2 is moving hor-

izontally, so it will take the same time to reach point C as fragment 1 (1.02

s).

Step 6: Calculate the distance CD. The horizontal distance CD is the speed

of fragment 2 multiplied by the time:

CD = 15 ×1.02

CD ≈15.3 m

Therefore, the distances are: AB = 10.7 m, BC = 5.1 m, and CD = 15.3 m.

Question 18

Question

A baseball player throws a ball with an initial velocity of 20 m/s at an angle of

45◦above the horizontal.

1. Determine the time the ball is in the air.

2. Calculate the maximum height the ball reaches.

(Note: Use g= 9.81 m/s2for the acceleration due to gravity)

Solution

1. To determine the time the ball is in the air, we can analyze the vertical

motion of the ball. The vertical component of the initial velocity can be found

using viy =visin(θ). Here, vi= 20 m/s and θ= 45◦.

viy = 20 sin(45◦) = 20 √2

2!= 10√2 m/s

The ball is thrown upwards, so the acceleration is in the negative ydirection.

We can use the equation vf=vi+at to find the time the ball is in the air. At

its highest point, the ball’s final velocity will be 0.

0 = 10√2−9.81t

t=10√2

9.81 ≈1.44 s

2. To find the maximum height the ball reaches, we can use the kinematic

equation yf=yi+viyt−1

2gt2. At the highest point, the final vertical displace-

ment is 0.

0 = 0 + 10√2(1.44) −1

2(9.81)(1.44)2

0 = 14.4−9.81(2.0736)

0 = 14.4−20.35

20.35 = 14.4 + ymax

ymax = 20.35 −14.4=5.95 m

Therefore, the time the ball is in the air is approximately 1.44 s, and the

maximum height the ball reaches is 5.95 m.

Question 19

Question

A baseball is hit with an initial velocity of 30 m/s at an angle of 45◦above the

horizontal. Calculate the maximum height reached by the baseball. Assume air

resistance is negligible and take the acceleration due to gravity to be −9.8 m/s2.

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The initial velocity can be broken down into its horizontal component (vix) and

vertical component (viy). These components can be found using trigonometric

functions:

vix=vicos(θ)

viy=visin(θ)

where vi= 30 m/s is the initial velocity and θ= 45◦is the angle above the

horizontal. Substitute the given values:

vix= 30 m/s ·cos(45◦)

viy= 30 m/s ·sin(45◦)

Solving for vixand viy:

vix= 30 m/s ·√2

2= 15√2 m/s

viy= 30 m/s ·√2

2= 15√2 m/s

Step 2: Calculate the time taken to reach the maximum height. The time

taken for the baseball to reach the maximum height can be found using the

formula:

vf=vi+at

where vf= 0 m/s is the final vertical velocity at the maximum height, vi=

15√2 m/s is the initial vertical velocity, and a=−9.8 m/s2is the acceleration

due to gravity. Substitute the given values:

0 = 15√2−9.8t

Solving for t:

t=15√2

9.8≈2.14 s

Step 3: Calculate the maximum height reached by the baseball. The maxi-

mum height (h) can be found using the formula:

h=viyt+1

2at2

Substitute the known values:

h= 15√2·2.14 −1

2·9.8·(2.14)2

Solving for h:

h≈21.4 m

Therefore, the maximum height reached by the baseball is approximately

21.4 meters.

Question 20

Question

A projectile is fired from the ground at an angle of 30◦above the horizontal with

an initial speed of 40 m/s. At the highest point in its trajectory, the projectile

explodes into two fragments. One fragment, with mass 2 kg, lands 40 m from

the point directly below the explosion 10 seconds later. Determine the mass of

the other fragment and the horizontal and vertical components of the velocity

of both fragments at the time of the explosion.

Solution

Step 1: Find the time to reach the highest point using the y-component of the

velocity. Given: Initial speed v0= 40 m/s, Angle of projection θ= 30◦,g= 9.8

m/s2.

The ycomponent of the initial velocity is:

viy=v0sin θ.

Using the kinematic equation vf=vi+at and vf= 0, we can solve for the time

to reach the highest point:

0 = viy+ (−g)t.

Solving for tgives:

t=viy

g=40 ×sin 30◦

9.8=40 ×0.5

9.8= 2.04 s.

Step 2: Find the vertical component of the fragment’s velocity at the highest

point. At the highest point, the vertical component of the velocity is zero. So,

we can find the ycomponent of the velocity using vfy=viy−gt:

vfy= 40 ×0.5−9.8×2.04 = 10.2−19.98 = −9.78 m/s.

Step 3: Find the horizontal component of the fragment’s velocity at the time

of the explosion. The horizontal component of the velocity remains constant

throughout the trajectory. So, the horizontal component of the velocity at the

time of explosion is the same as the initial horizontal component:

vfx=vix=v0cos θ= 40 ×cos 30◦= 40 ×0.866 = 34.64 m/s.

Step 4: Use the information to find the mass of the other fragment. Let

the mass of the other fragment be m. The horizontal distance it travels before

landing is 40 m. We can use the equation of motion

x=vixt

to find the time it takes for the other fragment to land:

40 = 34.64 ×t.

Thus,

t=40

34.64 = 1.155 s.

Step 5: Find the vertical component of the other fragment’s velocity at the

time of the explosion. Using vfy=viy−gt:

vfy= 40 ×0.5−9.8×1.155 = 10.2−11.31 = −1.11 m/s.

Therefore, the mass of the other fragment is 2 kg, and at the time of ex-

plosion, the horizontal component of its velocity is 34.64 m/s and the vertical

component is -1.11 m/s.

Question 21

Question

A baseball player throws a ball from the outfield towards home plate. The ball

leaves the player’s hand at an angle of 30◦above the horizontal with an initial

speed of 30 m/s. Given that the distance between the pitcher’s mound and

home plate is 20 meters, determine whether the ball will reach home plate on

the fly.

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

With the angle of 30◦, the initial velocity components are:

v0x=v0cos θ= 30 cos 30◦≈26.0 m/s

v0y=v0sin θ= 30 sin 30◦≈15.0 m/s

Step 2: Determine the time taken for the ball to reach the home plate using

the vertical motion equation:

y=v0yt−1

2gt2

0 = 15t−1

2(9.81)t2

The solution to this quadratic equation is t≈3.06 s.

Step 3: Calculate the horizontal distance the ball travels using the horizontal

motion equation:

x=v0xt

x= 26.0×3.06 ≈79.6 m

Since the distance between the pitcher’s mound and home plate is 20 meters,

the ball will sail over home plate and continue its trajectory beyond the field.

Question 22

Question

A projectile is launched from the ground at an angle of 30 degrees above the

horizontal with an initial speed of 20 m/s. Determine the time it takes for the

projectile to reach its maximum height.

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components.

The initial velocity can be broken down into its horizontal and vertical compo-

nents: The initial horizontal velocity (vix) is given by:

vix=vi·cos(θ)

vix= 20 m/s ·cos(30◦)

vix= 20 m/s ·√3

vix= 10√3 m/s

The initial vertical velocity (viy) is given by:

viy=vi·sin(θ)

viy= 20 m/s ·sin(30◦)

viy= 20 m/s ·1

viy= 10 m/s

Step 2: Determine the time to reach maximum height.

The time it takes for the projectile to reach its maximum height can be calcu-

lated using the vertical component of the velocity and the acceleration due to

gravity. The equation to use is:

vf=viy−g·t

where vfis the final vertical velocity (0 m/s at maximum height), viyis the

initial vertical velocity (10 m/s), gis the acceleration due to gravity (-9.8 m/s2),

and tis the time.

Solving for time:

0 = 10 −9.8t

9.8t= 10

t=10

9.8

t≈1.02 s

Therefore, it takes approximately 1.02 seconds for the projectile to reach its

maximum height.

Question 23

Question

A projectile is launched from ground level with an initial speed of 30 m/s at

an angle of 60◦above the horizontal. At the highest point of its trajectory,

the projectile explodes into two pieces of equal mass. One piece comes straight

down with no initial speed. Where does the other piece land relative to the

point directly below the point of explosion?

Solution

Step 1: Solve for the time of flight and the maximum height reached by the

projectile.

Let’s consider the horizontal and vertical components of the projectile mo-

tion separately.

In the vertical direction: - The projectile starts and ends at the same height

(y= 0 at both points). - The initial vertical velocity is viy= 30 sin 60◦= 15√3

m/s. - The vertical acceleration is ay=−9.8 m/s2(due to gravity).

Using the kinematic equation vf=vi+a·t, where vf= 0 at the highest

point of the trajectory:

0 = 15√3−9.8·t

Solving for the time t, we get:

t=15√3

9.8≈2.73 s

The time of flight is twice this value:

Time of flight = 2 ·2.73 = 5.46 s

The maximum height ymax reached can be calculated using the kinematic

equation ymax =viy·t+1

2ayt2:

ymax = 15√3·2.73 + 1

2·(−9.8) ·(2.73)2≈37.5 m

Step 2: Determine the horizontal distance traveled by the projectile.

The horizontal velocity remains constant at vix= 30 cos 60◦= 15 m/s.

The horizontal distance xtraveled can be found using the equation x=vix·t:

x= 15 ·5.46 ≈81.9 m

Therefore, the projectile lands 81.9 meters away horizontally from the launch

point.

Step 3: Determine the landing point of the other piece.

Since one piece comes straight down with no initial speed, it will hit the

ground directly below the point of explosion. Hence, the other piece will also

land at a horizontal distance of 81.9 meters from the point of explosion.

Question 24

Question

A baseball player hits a ball at an angle of 30◦above the horizontal with an

initial velocity of 40 m/s. The ball lands on the flat roof of a 10 m high building

100 m away. What is the time of flight of the ball?

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components. The

initial velocity can be decomposed into its horizontal and vertical components:

v0x=v0cos θ

v0y=v0sin θ

Substitute v0= 40 m/s and θ= 30◦into the above equations to get:

v0x= 40 cos 30◦= 40 ·√3

2= 20√3 m/s

v0y= 40 sin 30◦= 40 ·1

2= 20 m/s

Step 2: Calculate the time taken for the ball to reach the roof. The vertical

motion of the ball can be analyzed using the following kinematic equation:

y=v0yt+1

2ayt2

We know that y= 10 m (height of the building), v0y= 20 m/s, ay=−9.8

m/s2(acceleration due to gravity in the vertical direction), and we want to find

t. Plugging these values into the equation gives:

10 = 20t−1

2·9.8·t2

10 = 20t−4.9t2

4.9t2−20t+ 10 = 0

Solving this quadratic equation will give us the time taken for the ball to

reach the roof.

Question 25

Question

A projectile is launched from the ground at an angle of 30◦above the horizontal

with an initial speed of 20 m/s. At a certain instant, the projectile is at a height

of 15 m above the ground. Find the magnitude of the velocity of the projectile

at this instant.

Solution

Step 1: Break the initial velocity into horizontal and vertical components. The

initial velocity of the projectile is given by v0= 20 m/s and the launch angle

is θ= 30◦. The horizontal component of the velocity is v0x=v0cos θand the

vertical component of the velocity is v0y=v0sin θ.

Step 2: Find the time taken for the projectile to reach a height of 15 m.

The vertical motion of the projectile can be described by the equation y=

y0+v0yt−1

2gt2, where yis the height of the projectile, y0is the initial height,

v0yis the initial vertical velocity, gis the acceleration due to gravity, and tis

the time. We know that y0= 0 m, y= 15 m, v0y= 20 sin 30◦= 10 m/s, and

g= 9.8 m/s2. Plugging these values in, we get:

15 = 10t−1

2(9.8)t2

Solving for t, we get t≈1.53 s.

Step 3: Find the velocity of the projectile at this instant. At the instant the

projectile is at a height of 15 m, we know the time t= 1.53 s. The horizontal

velocity remains constant, so vx=v0x= 20 cos 30◦= 17.32 m/s. The vertical

velocity at this instant can be found using the equation vy=v0y−gt. Plugging

in the values, we get:

vy= 10 −9.8(1.53) = −4.79 m/s

The magnitude of the velocity at this instant is:

v=qv2

x+v2

y=p17.322+ (−4.79)2≈17.89 m/s

Therefore, the magnitude of the velocity of the projectile at this instant is

approximately 17.89 m/s.

Question 26

Question

A projectile is launched with an initial speed of 40 m/s at an angle of 30 degrees

above the horizontal. Find the maximum height reached by the projectile.

Solution

Step 1: Break the initial velocity into its xand ycomponents. The initial speed

of 40 m/s can be broken into its xand ycomponents:

vix=vicos θ= 40 cos 30◦= 40 ·√3

2≈34.64 m/s

viy=visin θ= 40 sin 30◦= 40 ·1

2= 20 m/s

Step 2: Calculate the time to reach maximum height. The time to reach

maximum height can be found using the ycomponent of the initial velocity and

acceleration due to gravity.

vf=vi+at

At the peak of the projectile’s motion, the final velocity is 0 m/s.

0 = 20 −9.8t

t=20

9.8≈2.04 s

Step 3: Find the maximum height reached. The maximum height can be

calculated using the formula h=viyt−1

2gt2.

h= 20 ×2.04 −1

2×9.8×(2.04)2

h≈20.48 m

Therefore, the maximum height reached by the projectile is approximately

20.48 meters.

Question 27

Question

A projectile is launched from the ground at an angle of 45◦above the horizontal.

The initial velocity of the projectile is 30 m/s. Calculate the maximum height

reached by the projectile. (Neglect air resistance and take g= 9.81 m/s2)

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The initial velocity v0can be resolved into horizontal component v0xand vertical

component v0yas follows:

v0x=v0cos θ= 30 cos 45◦

v0y=v0sin θ= 30 sin 45◦

Step 2: Determine the time taken to reach the maximum height. The time

taken to reach the maximum height can be determined using the formula:

vfy =v0y−gt

At the maximum height, the vertical component of the velocity will be 0. So,

we have:

0 = v0y−gtmax

Solving for tmax gives:

tmax =v0y

Step 3: Calculate the maximum height. The maximum height hmax can be

calculated using the formula:

hmax =v0ytmax −1

2gt2

max

Substitute the values of v0yand tmax into the equation to find hmax.

Question 28

Question

A projectile is launched at an angle of 30◦above the horizontal with an initial

speed of 20 m/s. Calculate the maximum height reached by the projectile.

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components.

The horizontal component is given by: vix=vi·cos(θ) The vertical component

is given by: viy=vi·sin(θ), where vi= 20 m/s and θ= 30◦. So, vix=

20 m/s ·cos(30◦)≈17.32 m/s, viy= 20 m/s ·sin(30◦)≈10 m/s.

Step 2: Calculate the time taken to reach the maximum height. Use the

equation: vfy=viy−gt, where vfy= 0 m/s (at the maximum height) and

g= 9.81 m/s2. So, 0 = 10 −9.81t, which gives t≈1.02 s.

Step 3: Calculate the maximum height reached. Use the equation: ymax =

yi+viyt−1

2gt2, where yi= 0 (initial vertical position) and t= 1.02 s. So,

ymax = 0 + 10 ·1.02 −1

2·9.81 ·(1.02)2≈5.10 m.

Therefore, the maximum height reached by the projectile is approximately

5.10 meters.

Question 29

Question

A projectile is launched from the ground with an initial speed of 30 m/s at an

angle of 30◦above the horizontal. At what time after being launched does the

projectile reach a height of 10 meters?

Solution

Step 1: Resolve the initial velocity into horizontal and vertical components. The

horizontal component of the initial velocity is given by v0x=v0cos θ, where v0

is the initial speed (30 m/s) and θis the angle of launch (30 degrees). Therefore,

v0x= 30 cos 30◦≈25.98 m/s

The vertical component of the initial velocity is given by v0y=v0sin θ, so

v0y= 30 sin 30◦= 15 m/s

Step 2: Determine the time it takes for the projectile to reach a height of 10

meters. Consider the motion of the projectile in the vertical direction. We will

use the kinematic equation:

y=y0+v0yt−1

2gt2

where y0is the initial height (0 meters), yis the final height (10 meters), v0yis

the initial vertical velocity (15 m/s), gis the acceleration due to gravity (-9.8

m/s2), and tis the time we need to find.

Plugging in the values we have:

10 = 0 + 15t−1

2(9.8)t2

Step 3: Solve for the time t. Rearranging the equation, we get:

4.9t2−15t+ 10 = 0

This is a quadratic equation that can be solved using the quadratic formula.

The solutions are:

t=15 ±p152−4(4.9)(10)

2(4.9)

t≈3.35 s or 0.60 s

Since the time cannot be negative, the projectile reaches a height of 10 meters

at approximately 0.60 seconds after being launched.

Question 30

Question

A ball is kicked with an initial velocity of 20 m/s at an angle of 30◦above the

horizontal. Find the maximum height reached by the ball.

Solution

Step 1: Break the initial velocity into its horizontal and vertical components.

The horizontal component, v0x, is given by v0x=v0cos(θ), where v0=

20 m/s and θ= 30◦. Therefore,

v0x= 20 m/s ×cos(30◦) = 20 m/s ×√3

2= 10√3 m/s

The vertical component, v0y, is given by v0y=v0sin(θ), where v0= 20 m/s

and θ= 30◦. Therefore,

v0y= 20 m/s ×sin(30◦) = 20 m/s ×1

2= 10 m/s

Step 2: Use the vertical component to find the time taken to reach the

maximum height.

The ball reaches its maximum height when the vertical velocity becomes 0.

We can use the equation vf=v0y−gt where vf= 0 and g= 9.8 m/s2.

0 = 10 −9.8t

t=10

9.8≈1.02 s

Step 3: Calculate the maximum height using the vertical component and

time.

The maximum height Hreached by the ball can be found using the equation

H=v0yt−1

2gt2.

H= 10 ×1.02 −1

2×9.8×(1.02)2

H≈5.1−5≈0.1 m

Therefore, the maximum height reached by the ball is approximately 0.1 m.

Question 31

Question

A projectile is launched from the ground at an angle of 30◦above the horizontal

with an initial speed of 20 m/s. At the highest point of the trajectory, the

projectile explodes into two fragments. One fragment, of mass 2 kg, moves

vertically upward with a speed of 4 m/s immediately after the explosion, while

the other fragment, of mass 3 kg, moves horizontally with a speed of 10 m/s.

Calculate the height above the ground at which the explosion occurs.

Solution

Step 1: Calculate the time of flight of the projectile motion.

The time of flight is given by:

T=2Visin θ

where Viis the initial velocity, θis the launch angle, and gis the acceleration

due to gravity.

Given Vi= 20 m/s, θ= 30◦, and g= 9.81 m/s2, we can calculate T:

T=2×20 ×sin(30◦)

9.81 ≈2.04 s

Step 2: Calculate the height at the top of the trajectory.

The maximum height Hreached by the projectile is given by:

H=V2

isin2θ

Plugging in the known values:

H=(20)2×(sin 30◦)2

2×9.81 ≈5.10 m

Step 3: Set up and solve simultaneous equations for the vertical and hori-

zontal components of the explosion.

Let hbe the height at which the explosion occurs. For the vertical motion,

we have:

0−h= 4t−1

2gt2

where tis the time taken for the projectile to reach height h.

For the horizontal motion, we have:

0−0 = 10t

The times for both the vertical and horizontal motion are the same, so we can

solve for tfrom the horizontal motion equation:

t=h

Substitute this time into the vertical motion equation:

0−h= 4 ×h

10 −1

2gh

102

Solving this equation gives the height of the explosion, h:

h= 15.6 m

Therefore, the height above the ground at which the explosion occurs is 15.6

Question 32

Question

A particle moves along a curved path in the xy plane according to the following

equations: x(t) = 3t2−2tand y(t) = 4t3−t. Find the particle’s velocity and

acceleration vectors at t= 2 seconds.

Solution

Step 1: To find the particle’s velocity, we first calculate the derivatives of the

position functions x(t) and y(t) to obtain the x(t) and y(t) components of ve-

locity.

Given x(t)=3t2−2tand y(t) = 4t3−t, the velocity components are:

vx(t) = dx

dt =d

dt (3t2−2t)=6t−2

vy(t) = dy

dt =d

dt (4t3−t) = 12t2−1

At t= 2 s, the velocity vector is:

v=vx(2)i+vy(2)j= 10i+ 47jm/s

Step 2: Next, we find the particle’s acceleration vector by taking the deriva-

tives of the velocity components.

The acceleration components are:

ax(t) = dvx

dt =d

dt (6t−2) = 6

ay(t) = dvy

dt =d

dt (12t2−1) = 24t

At t= 2 s, the acceleration vector is:

a=ax(2)i+ay(2)j= 6i+ 48jm/s2

Question 33

Question

A particle starts at the origin of an xy-coordinate system and moves in the

xy-plane with velocity components vx= 3tm/s and vy= 4 −2tm/s, where tis

in seconds. Determine the magnitude and direction of the particle’s velocity at

t= 2 s.

Solution

Step 1: Recall that the velocity of a particle in two dimensions is given by

v =vxˆ

i+vyˆ

j, where vxis the velocity component in the x-direction, and vyis

the velocity component in the y-direction.

Given vx= 3tm/s and vy= 4 −2tm/s, we have v = 3tˆ

i+ (4 −2t)ˆ

Step 2: To find the magnitude of the velocity, we use the formula |v|=

qv2

x+v2

Substitute the given values at t= 2 s:

|v|=p(3(2))2+ (4 −2(2))2

|v|=√36 + 0

|v|=√36

|v|= 6 m/s

Step 3: To find the direction of the velocity, we use the formula θ=

tan−1vy

vx.

Substitute the given values at t= 2 s:

θ= tan−14−2(2)

3(2) 

θ= tan−10

6

θ= tan−1(0)

θ= 0◦

Therefore, at t= 2 s, the magnitude of the particle’s velocity is 6 m/s, and

its direction is in the positive x-direction.

Question 34

Question

A baseball is hit into the outfield at an angle of 35◦above the horizontal with

an initial speed of 30 m/s. How far from home plate does the baseball land if it

is caught by an outfielder 1.5 m above the ground?

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The initial velocity of the baseball can be resolved into its horizontal and vertical

components as follows:

V0x=V0cos θ= 30 cos 35◦≈24.69 m/s

V0y=V0sin θ= 30 sin 35◦≈17.21 m/s

Step 2: Calculate the time it takes for the baseball to reach the outfielder.

The time taken for the baseball to reach the outfielder can be calculated using

the vertical component of the motion with the equation:

y=V0yt+1

2gt2

where y= 1.5 m (height of the outfielder) and g= 9.81 m/s2(acceleration due

to gravity). Solving for t:

1.5 = 17.21t−4.905t2

4.905t2−17.21t+ 1.5=0

Using the quadratic formula, we find the positive value for tto be approximately

t≈1.42 s.

Step 3: Calculate the horizontal distance the baseball travels before reaching

the outfielder. The horizontal distance can be calculated using the equation:

x=V0xt

x= 24.69 ×1.42 ≈35.06 m

Therefore, the baseball lands approximately 35.06 meters away from home

plate.

Question 35

Question

A baseball is hit at an angle of 35◦above the horizontal with an initial speed

of 30 m/s from a height of 2 m above the ground. Ignoring air resistance,

determine the maximum height the baseball reaches and the total time it’s in

the air.

Solution

Step 1: Resolve the initial velocity into its horizontal and vertical components.

The initial velocity of the baseball can be resolved into horizontal and vertical

components as follows:

vix=vicos(θ) = 30 cos(35◦)≈24.63 m/s

viy=visin(θ) = 30 sin(35◦)≈17.14 m/s

Step 2: Determine the time it takes to reach the maximum height.

To find the time it takes to reach the maximum height, we use the vertical

equation of motion:

y=yi+viyt−1

2gt2

where yi= 2 m is the initial height, yis the maximum height (which occurs

when the vertical velocity becomes 0), and g= 9.81 m/s2is the acceleration

due to gravity.

At the maximum height, vy= 0, so:

0 = 17.14 −9.81t

Solving for tgives:

t=17.14

9.81 ≈1.75 s

Step 3: Calculate the maximum height reached by the baseball.

Substitute the time found in Step 2 into the equation for vertical position

with viy= 17.14 m/s:

y= 2 + 17.14 ×1.75 −1

2×9.81 ×(1.75)2

y≈24.86 m

Step 4: Determine the time the baseball is in the air.

The total time the baseball is in the air is twice the time it takes to reach

the maximum height (up and back down again):

Total time in air = 2 ×1.75 = 3.50 s

Therefore, the maximum height the baseball reaches is approximately 24.86

m and the total time it is in the air is 3.50 seconds.

PHYS 202 - GENERAL PHYSICS II Kinematics in Two Dimensions Question Bank Set 2

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