PHYS 202 - GENERAL PHYSICS II Kinematics in Two Dimensions Question Bank Set 1

PHYS 202 - GENERAL PHYSICS II -

Kinematics in Two Dimensions

Question Bank - Set 1

Liberty University

Question 1

Question

A particle moves in a plane in such a way that its position vector at time tis

given by r(t) = (3t2−2)ˆ

i+ (4t+ 1)ˆ

j, where ˆ

iand ˆ

jare unit vectors in the

positive xand ydirections, respectively. Find the magnitude of the particle’s

velocity and acceleration at t= 2 s.

Solution

Step 1: To find the velocity of the particle, we take the derivative of the position

vector with respect to time.

Velocity, v(t) = dr

dt =d

dt[(3t2−2)ˆ

i+ (4t+ 1)ˆ

v(t) = (6t)ˆ

i+ 4ˆ

Step 2: Now, we can find the velocity of the particle at t= 2 s.

v(2) = (6(2))ˆ

i+ 4ˆ

j= 12ˆ

i+ 4ˆ

Step 3: To find the magnitude of the velocity, we calculate |v(2)|.

|v(2)|=p(12)2+ (4)2=√144 + 16 = √160 = 4√10 m/s

Step 4: Next, to find the acceleration of the particle, we take the derivative

of the velocity vector with respect to time.

Acceleration, a(t) = dv

dt =d

dt[(6t)ˆ

i+ 4ˆ

a(t)=6ˆ

Step 5: Now, we can find the acceleration of the particle at t= 2 s.

a(2) = 6ˆ

Step 6: To find the magnitude of the acceleration, we calculate |a(2)|.

|a(2)|=p(6)2= 6 m/s2

Therefore, at t= 2 s, the magnitude of the particle’s velocity is 4√10 m/s

and the magnitude of the acceleration is 6 m/s2.

Question 2

Question

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

speed of 50 m/s. Calculate the maximum height reached by the projectile.

Solution

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

The initial velocity can be broken into horizontal and vertical components as

follows: Vix=Vi·cos(30◦) = 50 m/s ·cos(30◦)Viy=Vi·sin(30◦) = 50 m/s ·

sin(30◦)

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

taken for the projectile to reach the maximum height can be determined using

the vertical component of the initial velocity and the acceleration due to gravity:

Vfy= 0 m/s (at maximum height) Using the kinematic equation Vfy=Viy−gt,

where g= 9.81 m/s2: 0 = 50 m/s ·sin(30◦)−9.81 m/s2·tSolving for tgives us

the time taken to reach the maximum height.

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

using the vertical motion equation: Hmax =Viy·t−1

2gt2Substitute the values

of Viyand tinto the equation to find the maximum height.

Question 3

Question

A soccer player kicks a ball from ground level with an initial velocity of 20 m/s

at an angle of 45◦above the horizontal. Calculate the maximum height the ball

reaches and the total time of flight.

Solution

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

The initial velocity vi= 20 m/s can be resolved into horizontal and vertical

components as follows: Horizontal component: vix=vicos(45◦) = 20 ·√2

10√2 m/s Vertical component: viy=visin(45◦) = 20 ·√2

2= 10√2 m/s

Step 2: Calculate the time it takes for the ball to reach the maximum height.

Using the vertical motion equation vf=vi+at where vf= 0 m/s (at the

maximum height) and a=−9.81 m/s2(acceleration due to gravity), we have:

0 = 10√2−9.81tSolving for t, we get: t=10√2

9.81 s

Step 3: Calculate the maximum height the ball reaches. Using the vertical

motion equation y=vit+1

2at2with vi= 10√2 m/s, a=−9.81 m/s2and t

from Step 2, we have: y= 10√2·10√2

9.81 +1

2(−9.81) 10√2

9.81 2Solving for y, we

get: y=(200)

9.81 −(200)

9.81 =200

9.81 m

Step 4: Calculate the total time of flight. Since the motion is symmetrical,

the time of flight is twice the time it takes to reach the maximum height: Total

time of flight = 2 ·10√2

9.81 s

Question 4

Question

A projectile is launched from the ground with an initial speed of 40 m/s at an

angle of 30 degrees above the horizontal. Find the maximum height above the

ground reached by the projectile. (Neglect air resistance and take g= 9.8 m/s2)

Solution

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

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

components as follows:

v0x=v0cos(θ) = 40 cos(30◦)≈34.64 m/s

v0y=v0sin(θ) = 40 sin(30◦)≈20 m/s

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

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

the equation of motion: vf=vi+at, where vf= 0, vi= 20 m/s, a=−9.8

m/s2, and tis the time taken to reach the maximum height, we have:

0 = 20 −9.8t

t=20

9.8≈2.04 s

Step 3: Calculate the maximum height reached by the projectile. Using the

equation of motion: s=vit+1

2at2, where sis the displacement in the vertical

direction, vi= 20 m/s, a=−9.8 m/s2, and t= 2.04 s, we have:

s= 20(2.04) + 1

2(−9.8)(2.04)2

s≈20.4 m

Therefore, the maximum height above the ground reached by the projectile

is approximately 20.4 meters.

Question 5

Question

A quarterback throws a football at an angle of 45◦above the horizontal with

an initial speed of 20 m/s. How far away does the football land from the

quarterback?

Solution

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

The initial velocity can be broken down into its horizontal component (vi,x)

and vertical component (vi,y ) using trigonometry. Since the angle is 45◦, both

components are equal.

vi,x =vi,y =vi·cos 45◦=vi·sin 45◦= 20 m/s ·1

√2≈14.14 m/s

Step 2: Determine the time of flight. The time taken for the football to

reach the highest point can be found using the vertical component of the initial

velocity and acceleration due to gravity.

vf,y =vi,y −g·t

0 = 14.14 m/s −9.81 m/s2·t

t=14.14 m/s

9.81 m/s2≈1.44 s

Step 3: Calculate the horizontal distance. The horizontal distance can be

calculated using the horizontal component of the initial velocity and the time

of flight.

Distance = vi,x ·t

Distance = 14.14 m/s ·1.44 s ≈20.36 m

Therefore, the football lands approximately 20.36 meters away from the

quarterback.

Question 6

Question

A baseball player hits a ball with an initial velocity of 40 m/s at an angle of

30◦above the horizontal. The ball lands on the ground 150 meters away. How

long is the ball in the air? What is the maximum height the ball reaches?

Solution

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

The initial velocity of the ball can be resolved into its horizontal (v0x) and

vertical (v0y) components using trigonometry:

v0x=v0cos(θ)

v0y=v0sin(θ)

Where v0= 40 m/s and θ= 30◦.

Substitute the values to find v0xand v0y:

v0x= 40 cos(30◦) = 34.64 m/s

v0y= 40 sin(30◦) = 20 m/s

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

the equation for time taken for an object to fall freely from an initial height to

the ground:

t=2v0y

Where g= 9.81 m/s2is the acceleration due to gravity.

Substitute v0yinto the equation to find t:

t=2×20

9.81 = 4.08 s

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

the ball can be determined using the equation for vertical motion:

h=v2

0y/(2g)

Substitute the values of v0yand gto calculate the maximum height:

h= (202)/(2 ×9.81) = 20.41 m

Therefore, the ball is in the air for 4.08 seconds and reaches a maximum

height of 20.41 meters.

Question 7

Question

A baseball is hit with an initial speed of 30 m/s at an angle of 30 degrees above

the horizontal. How far does the baseball travel horizontally before landing on

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

Solution

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

The initial velocity v0can be resolved into its horizontal component v0xand

vertical component v0yas follows:

v0x=v0cos θ

v0y=v0sin θ

where v0= 30 m/s and θ= 30◦. Plugging in the values, we get:

v0x= 30 m/s ×cos 30◦= 26 m/s

v0y= 30 m/s ×sin 30◦= 15 m/s

Step 2: Determine the time of flight (t). The time of flight for the baseball

can be found using the formula for vertical motion:

∆y=v0yt+1

2at2

where ∆yis the vertical displacement (equal to the height the baseball is hit

from, which we’ll assume is 0), a=−g(negative because it is acting in the

downward direction). Rearranging for t, we have:

0 = 15t−1

2×9.8×t2

Solving this quadratic equation gives t= 3.06 s.

Step 3: Calculate the horizontal distance traveled (∆x). The horizontal

distance ∆xcan be found using the formula for horizontal motion:

∆x=v0x×t

Plugging in the values, we get:

∆x= 26 m/s ×3.06 s = 79.56 m

Therefore, the baseball travels approximately 79.56 meters horizontally be-

fore landing on the ground.

Question 8

Question

A ball is thrown horizontally from the top of a building that is 50 meters high.

The initial velocity of the ball is 20 m/s. How far from the base of the building

does the ball land?

Solution

Step 1: Identify the known variables. The initial vertical velocity of the ball

is 0 m/s (thrown horizontally). The final vertical position of the ball is 50 m

(height of the building). The acceleration in the vertical direction is due to

gravity, which is −9.8 m/s2. The initial horizontal velocity of the ball is 20 m/s

and the ball lands on the ground at some horizontal distance we need to find.

Step 2: Determine the time taken for the ball to reach the ground. Since

the ball is only influenced by gravity in the vertical direction, we can use the

equation:

yf=yi+vyit+1

2at2.

Plug in the known values:

50 m = 0 m + 0 m/s ·t+1

2·(−9.8 m/s2)·t2.

Solving for t, we get:

t=s2·50 m

9.8 m/s2≈3.20 s.

Step 3: Calculate the horizontal distance traveled by the ball. Since the

horizontal velocity of the ball is constant, we can use the equation:

xf=xi+vxit.

Plug in the known values:

xf= 0 m + 20 m/s ·3.20 s = 64 m.

Answer: The ball lands 64 meters from the base of the building.

Question 9

Question

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

The initial speed of the cannonball is 50 m/s. Calculate the following:

1. The maximum height the cannonball reaches.

2. The time it takes for the cannonball to reach the maximum height.

3. The total time the cannonball is in the air before hitting the ground.

(Assume no air resistance.)

Solution

Let’s denote the initial speed of the cannonball as v0= 50 m/s, the launch angle

as θ= 30◦, the acceleration due to gravity as g= 9.81 m/s2, and the maximum

height reached by the cannonball as H. We will use the kinematic equations to

solve for the different parts of the question.

Part 1: Maximum height reached by the cannonball

Step 1: We can determine the vertical component of the initial velocity

(v0y) using the initial speed and launch angle:

v0y=v0sin(θ)

v0y= 50 m/s ·sin(30◦)

v0y≈25 m/s

Step 2: Next, we can use the kinematic equation for vertical motion to find

the maximum height:

fy=v2

0y−2·g·H

At the maximum height, vfy= 0.

0 = (25 m/s)2−2·9.81 m/s2·H

H=(25 m/s)2

2·9.81 m/s2

H≈31.93 m

Thus, the maximum height reached by the cannonball is approximately 31.93

meters.

Part 2: Time to reach the maximum height

Step 1: We can find the time it takes for the cannonball to reach the

maximum height using the vertical component of the initial velocity:

vfy=v0y−g·t

At the maximum height, vfy= 0.

0 = 25 m/s −9.81 m/s2·t

t=25 m/s

9.81 m/s2

t≈2.55 s

Therefore, it takes approximately 2.55 seconds for the cannonball to reach

the maximum height.

Part 3: Total time in the air

Since the motion is symmetric, the total time the cannonball is in the air

before hitting the ground is twice the time it takes to reach the maximum height.

Total time = 2 ×(Time to reach maximum height)

Total time = 2 ×2.55 s

Total time ≈5.10 s

Therefore, the total time the cannonball is in the air before hitting the

ground is approximately 5.10 seconds.

Question 10

Question

A baseball is hit with an initial velocity of 30 m/s in the horizontal direction

and 20 m/s in the vertical direction. Calculate the baseball’s maximum height,

the time it takes to reach the highest point, and the total time of flight. Assume

air resistance is negligible and take the acceleration due to gravity as 9.8 m/s2.

Solution

Step 1: Find the time to reach the highest point. Given that the initial ver-

tical velocity of the baseball is 20 m/s and the acceleration due to gravity is

−9.8 m/s2, we can use the kinematic equation:

vf=vi+at

where vfis the final velocity (0 m/s at the highest point), viis the initial velocity

(20 m/s up), ais the acceleration due to gravity (-9.8 m/s2), and tis the time

taken to reach the highest point. Substitute the known values:

0 = 20 −9.8t

Solving for t:

t=20

9.8≈2.04 s

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

for vertical motion:

yf=yi+vit+1

2at2

where yfis the final height (maximum height), yiis the initial height (0 m), vi

is the initial vertical velocity (20 m/s), ais the acceleration due to gravity (-9.8

m/s2), and tis the time taken to reach the highest point (2.04 s). Substitute

the known values:

yf= 0 + 20 ×2.04 −1

2×9.8×(2.04)2

yf≈20 ×2.04 −1

2×9.8×4.1616

yf≈40.8−20.4096

yf≈20.3904 m

Step 3: Determine the total time of flight. The total time of flight is twice

the time taken to reach the highest point.

T otal time = 2 ×2.04

T otal time = 4.08 s

Therefore, the baseball’s maximum height is 20.39 m, the time it takes to

reach the highest point is approximately 2.04 s, and the total time of flight is

4.08 s.

Question 11

Question

A soccer player kicks a ball from the ground with an initial speed of 20 m/s at

an angle of 30 degrees above the horizontal. The ball lands on the ground 4

seconds later. (a) What is the horizontal distance the ball travels? (b) What is

the maximum height the ball reaches? (c) What are the horizontal and vertical

components of the ball’s velocity when it lands?

Solution

Step 1: Determine the horizontal distance the ball travels.

Given: v0= 20 m/s, θ= 30◦,t= 4 s

The horizontal component of the initial velocity is v0x=v0cos(θ) and the

vertical component of the initial velocity is v0y=v0sin(θ).

Using the horizontal distance formula: x=v0x·t, we can find x.

v0x= 20 m/s ·cos(30◦) = 20 m/s ·√3

2= 10√3 m/s

x= 10√3 m/s ·4 s = 40√3 m ≈69.3 m

Therefore, the horizontal distance the ball travels is approximately 69.3 me-

ters.

Step 2: Determine the maximum height the ball reaches.

Using the vertical distance formula: y=v0yt−1

2gt2, where g= 9.8 m/s2.

Substitute the given values into the formula to find the maximum height,

ymax.

v0y= 20 m/s ·sin(30◦) = 20 m/s ·1

2= 10 m/s

ymax = 10 m/s ·4 s −1

2·9.8 m/s2·(4 s)2

ymax = 40 m −78.4 m = −38.4 m

The negative sign indicates the ball is below the initial point, so the max-

imum height the ball reaches is approximately 38.4 meters below the initial

point.

Step 3: Find the horizontal and vertical components of the ball’s velocity

when it lands.

The vertical component of the velocity when the ball lands is the same as the

initial vertical component due to symmetry. Therefore, the vertical component

is vfinaly=−10 m/s.

Since there is no horizontal acceleration, the horizontal component of the

velocity remains constant. Therefore, the horizontal component is vfinalx=

10√3 m/s.

Therefore, the horizontal and vertical components of the ball’s velocity when

it lands are 10√3 m/s and −10 m/s respectively.

Question 12

Question

A projectile is launched with an initial velocity of 30 m/s at an angle of 45◦

above the horizontal. Determine the maximum height the projectile reaches.

Given: g= 9.81 m/s2

Solution

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

The initial velocity, vi, can be resolved into horizontal and vertical components

using trigonometric functions. The vertical component is:

viy=visin θ

viy= 30 sin 45◦

viy≈21.21 m/s

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

to reach maximum height (tmax) can be found using the vertical component of

the initial velocity and the acceleration due to gravity. At maximum height, the

vertical component of velocity becomes 0 m/s.

vfy= 0 m/s = viy−g·tmax

0 = 21.21 −9.81 ·tmax

tmax ≈2.16 s

Step 3: Calculate the maximum height. The maximum height (Hmax) can

be determined using the kinematic equation for vertical motion:

Hmax =viy·tmax −1

2g·t2

max

Hmax = 21.21 ·2.16 −1

2·9.81 ·(2.16)2

Hmax ≈23.10 m

Therefore, the maximum height the projectile reaches is approximately 23.10

meters.

Question 13

Question

A projectile is launched from ground level at an angle of 60◦above the horizontal

with an initial speed of 20 m/s.

(a) Determine the maximum height reached by the projectile.

(b) Find the total time of flight.

Solution

(a) To find the maximum height reached by the projectile, we need to determine

the vertical component of the initial velocity.

Step 1: Find the initial vertical velocity component (viy). Given that the

initial speed is 20 m/s and the launch angle is 60◦, we have:

viy=visin(θ) = 20 m/s ×sin(60◦).

Therefore, viy= 20 ×√3

2≈17.32 m/s.

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

height, the vertical velocity component is 0 m/s. Using the equation vf=vi+at,

we have:

0 = 17.32 −9.81t.

Solving for tgives us:

t=17.32

9.81 ≈1.77 s.

Step 3: Find the maximum height. The maximum height (hmax) can be

determined using the equation:

hmax =viyt−1

2gt2.

Substitute the known values to get:

hmax = 17.32 ×1.77 −1

2×9.81 ×(1.77)2.

Calculating this gives hmax ≈15.15 m.

Therefore, the maximum height reached by the projectile is approximately

15.15 meters.

(b) To find the total time of flight, we can use the fact that the time to reach

maximum height is half the total time of flight.

The total time of flight is:

Total time = 2 ×1.77 ≈3.54 s.

formula:

Range = vix×Total time of flight,

where vixis the initial horizontal velocity component. This can be calculated

as:

vix=vicos(θ) = 20 m/s ×cos(60◦).

Thus, vix= 20 ×1

2= 10 m/s.

Substitute the values to get:

Range = 10 ×3.54 = 35.4 m.

Therefore, the horizontal range of the projectile is 35.4 meters.

Question 14

Question

An object is launched from ground level with an initial velocity of 20 m/s at an

angle of 30◦above the horizontal. Find the maximum height reached by the

object.

Solution

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

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

components as follows:

Vix=Vicos θ= 20 m/s cos 30◦

Viy=Visin θ= 20 m/s sin 30◦

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

to reach maximum height can be found using the equation v=vi+at. In

the vertical direction, the final velocity at maximum height is 0 m/s, the initial

velocity is Viy, the acceleration is due to gravity (−9.8 m/s2), and the time is

unknown.

0 = Viy−9.8 m/s2·t

Solving for t, we get:

t=Viy

9.8 m/s2

Step 3: Find the maximum height reached by the object. The maximum

height reached can be calculated using the equation h=Viyt−1

2gt2, where his

the maximum height.

h=Viy·Viy

9.8 m/s2−1

2·(−9.8 m/s2)· Viy

9.8 m/s2!2

Simplify the expression to find the maximum height reached by the object.

Question 15

Question

A projectile is launched at an angle of 45◦above the horizontal with an initial

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

Solution

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

Let v0= 20 m/s be the initial speed. The horizontal component of the initial

velocity v0x=v0cos(45◦) and the vertical component of the initial velocity

v0y=v0sin(45◦).

Step 2: Calculate the time to reach the maximum height. Consider the ver-

tical motion of the projectile. At the maximum height, the vertical component

of the velocity will be zero. Use the equation vyf =v0y−gt where vyf = 0 m/s.

Solve for tto find the time to reach the maximum height.

Step 3: Calculate the maximum height. Using the vertical motion equation

y=y0+v0yt−1

2gt2, where y0= 0 m (starting from the ground) and the time

tcalculated in Step 2, find the maximum height yreached by the projectile.

Step 4: Substitute the known values and solve for the maximum height.

Substitute v0y,t, and g= 9.81 m/s2into the equation from Step 3 to find the

maximum height y.

Step 1:

The horizontal component of the initial velocity is given by:

v0x=v0cos(45◦) = 20 ×1

√2= 10√2 m/s

The vertical component of the initial velocity is given by:

v0y=v0sin(45◦) = 20 ×1

√2= 10√2 m/s

Step 2:

Using the equation vyf =v0y−gt and rearranging for t, we get:

0 = 10√2−9.81t

t=10√2

9.81 ≈1.43 s

Step 3:

Substitute y0= 0 m, v0y= 10√2 m/s, t= 1.43 s, and g= 9.81 m/s2into the

vertical motion equation:

y= 0 + 10√2×1.43 −1

2×9.81 ×(1.43)2

Step 4:

Calculating the maximum height:

y≈10.2 m

Therefore, the maximum height reached by the projectile is approximately

10.2 meters.

Question 16

Question

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

with an initial speed of 20 m/s. What are the maximum height above the ground

and the range of the projectile?

Solution

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

The initial velocity v0= 20 m/s can be resolved into its horizontal and vertical

components as: Horizontal component: v0x=v0cos 45◦Vertical component:

v0y=v0sin 45◦

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

using the vertical motion equation: y=v0yt−1

2gt2where yis the height,

g= 9.81 m/s2is the acceleration due to gravity, and the projectile reaches its

maximum height when vy= 0. Solving for time t, we get: 0 = v0yt−1

2gt2

t=2v0y

Step 3: Calculate the maximum height reached by the projectile. The max-

imum height hreached by the projectile can be calculated using the vertical

motion equation: h=v0yt−1

2gt2

Step 4: Calculate the range of the projectile. The range Rof the projectile

can be calculated using the horizontal motion equation: R=v0x×2t

Step 5: Substitute known values and calculate. Now, substitute the given

values (v0= 20 m/s, g= 9.81 m/s2,v0y= 20 ×sin 45◦,v0x= 20 ×cos 45◦) into

the formulas found in Steps 2, 3, and 4 to calculate the maximum height and

range of the projectile.

Question 17

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 of its trajectory, the projectile

explodes into two fragments of equal mass. One fragment, whose initial speed

is negligible, falls vertically from the highest point. Neglecting air resistance,

what is the speed of the other fragment just after the explosion?

Solution

Step 1: Determine the components of the initial velocity of the projectile. Given

that the initial speed of the projectile is 40 m/s and it is fired at an angle of 30◦

above the horizontal, we can find the initial xand ycomponents of the velocity

as follows:

v0x=v0cos(θ) = 40 cos(30◦) = 40 ·√3

2= 20√3 m/s

v0y=v0sin(θ) = 40 sin(30◦) = 40 ·1

2= 20 m/s

Step 2: Determine the time taken for the projectile to reach the highest

point. Since the vertical component of the initial velocity is zero at the highest

point, we can use the equation:

vy=v0y−gt

where vy= 0 and g= 9.81 m/s2(acceleration due to gravity). Solving for t, we

get:

0 = 20 −9.81t=⇒t=20

9.81 ≈2.04 s

Step 3: Find the maximum height reached by the projectile. Using the

formula for vertical displacement with constant acceleration:

y=v0yt−1

2gt2

Plugging in the values, we get:

y= 20 ·2.04 −1

2·9.81 ·(2.04)2= 40.8−1

2·9.81 ·4.1616 ≈40.8−20.48 ≈20.32 m

Step 4: Determine the velocity of the fragment falling vertically. Since the

other fragment falls vertically from the highest point, its velocity just after the

explosion is equal to the velocity of the ball at the highest point. Thus, its

velocity is given by the vertical component of the initial velocity:

v′

0y= 20 m/s

Step 5: Calculate the speed of the other fragment just after the explosion.

The speed of a vector is given by the magnitude of the vector. Therefore, the

speed of the other fragment just after the explosion is:

v′=q(v0x)2+ (v′

0y)2=q(20√3)2+ (20)2=√1200 + 400 = √1600 = 40 m/s

Therefore, the speed of the other fragment just after the explosion is 40 m/s.

Question 18

Question

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

with an initial speed of 40 m/s. Determine the maximum height the projectile

reaches during its flight.

Solution

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

The vertical component can be found using viy=visin θand the horizontal

component can be found using vix=vicos θ.

Given: Initial speed, vi= 40 m/s Angle, θ= 30◦

Therefore: viy= 40 m/s ×sin(30◦)≈20 m/s vix= 40 m/s ×cos(30◦)≈

34.64 m/s

Step 2: Calculate the time when the projectile reaches its maximum height

using the equation vf=vi+at for vertical motion, where the final vertical

velocity will be zero at the peak.

Given: Final vertical velocity, vfy= 0 Initial vertical velocity, viy= 20 m/s

Vertical acceleration, ay=−9.8 m/s2

Substitute the values into the equation and solve for t: 0 = 20 m/s −

9.8 m/s2×t

t=20 m/s

9.8 m/s2≈2.04 s

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

motion and the time calculated in Step 2.

Use the equation y=vit+1

2at2with vertical displacement y=?, initial

vertical velocity viy= 20 m/s, vertical acceleration ay=−9.8 m/s2, and time

t= 2.04 s.

Substitute the values into the equation: y= 20 m/s×2.04 s+ 1

2(−9.8 m/s2)(2.04 s)2

y≈20.40 m + (−20.00 m) ≈0.40 m

Therefore, the projectile reaches a maximum height of approximately 0.40

meters during its flight.

Question 19

Question

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

with an initial speed of 30 m/s. Find the maximum height reached by the

projectile during its flight.

Solution

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

The vertical component is given by:

V0y=V0·sin θ

V0y= 30 m/s ·sin 45◦≈21.21 m/s

Step 2: Use the kinematic equation for vertical motion to find the time it

takes for the projectile to reach its maximum height. At maximum height, the

vertical velocity is 0 m/s.

vy=v0y−gt

0 = 21.21 m/s −9.8 m/s2·t

t≈2.17 s

Step 3: Use the kinematic equation for vertical motion to find the maximum

height reached by the projectile.

y=v0yt−1

2gt2

y= 21.21 m/s ·2.17 s −1

2·9.8 m/s2·(2.17 s)2

y≈23.57 m

Therefore, the maximum height reached by the projectile during its flight is

approximately 23.57 meters.

Question 20

Question

A projectile is fired from the ground with an initial speed of 80 m/s at an angle

of 30 degrees 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 (vix) and vertical (viy)

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

components as follows:

vix=vi·cos(θ)

viy=vi·sin(θ)

where vi= 80 m/s and θ= 30◦. Thus,

vix= 80 ·cos(30◦) = 80 ·√3

2= 40√3 m/s

viy= 80 ·sin(30◦) = 80 ·1

2= 40 m/s

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

imum height, the vertical component of the projectile’s velocity becomes zero.

Using the vertical motion equation vf=vi+at, where vf= 0 and a=−9.8 m/s2

(acceleration due to gravity), we can solve for time t:

0 = 40 −9.8t

t=40

9.8≈4.08 s

Therefore, it takes approximately 4.08 seconds for the projectile to reach its

maximum height.

Question 21

Question

A projectile is fired at an angle of 45◦above the horizontal with an initial speed

of 50 m/s. Find the maximum height reached by the projectile.

Solution

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

The initial velocity v0can be decomposed into its horizontal component (v0x)

and vertical component (v0y) as follows: v0x=v0cos(45◦) and v0y=v0sin(45◦).

Given that v0= 50 m/s, the initial horizontal and vertical components of

the velocity are: v0cos(45◦) = 50 cos(45◦) and v0sin(45◦) = 50 sin(45◦).

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

mum height, the vertical component of the projectile’s velocity is zero. Use the

kinematic equation vy=v0y−gt and solve for the time twhen vy= 0.

0 = v0y−gt =⇒t=v0y

g, where g= 9.81 m/s2is the acceleration due to

gravity.

Substitute v0y= 50 sin(45◦) into the equation to find the time t.

Step 3: Calculate the maximum height reached by the projectile. The max-

imum height can be calculated using the equation for vertical displacement:

ymax =v0yt−1

2gt2.

Substitute v0y= 50 sin(45◦) and t=50 sin(45◦)

9.81 into the equation to find the

maximum height ymax.

Question 22

Question

A baseball player hits a baseball such that it reaches a maximum height of 20

meters and lands 100 meters away. Assuming the baseball is hit from a height

of 1 meter above the ground and neglecting air resistance, what was the initial

speed of the baseball?

Solution

Step 1: Determine the time taken for the baseball to reach the maximum height.

The vertical motion of the baseball can be analyzed using the kinematic equa-

tion:

y=viyt−1

2gt2

where: y= maximum height = 20 m, viy= initial vertical velocity, g= accel-

eration due to gravity = 9.81 m/s2.

At the maximum height, the vertical velocity is 0. Therefore,

0 = viy−gtmax

Solving for tmax:

tmax =viy

Using this time to find the initial vertical velocity:

20 = viyviy

g−1

2gviy

g2

Solve for viy.

Step 2: Determine the horizontal speed of the baseball. In the absence of

air resistance, there is no horizontal acceleration. Thus, the horizontal speed

remains constant throughout the motion. Using the horizontal distance x= 100

m and time of flight ttotal = 2tmax:

vix=x

ttotal

Step 3: Use the horizontal and vertical speeds to find the initial speed of the

baseball. The initial speed can be calculated using the Pythagorean theorem:

vi=qv2

ix+v2

Substitute the values of vixand viyto find the initial speed vi.

Question 23

Question

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

the horizontal. Find the maximum height the projectile reaches during its flight.

Solution

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

initial velocity v0= 30 m/s can be resolved into horizontal (v0x) and vertical

components (v0y) as follows:

v0x=v0cos(60◦) and v0y=v0sin(60◦)

v0x= 30 cos(60◦)≈15 m/s and v0y= 30 sin(60◦)≈25.98 m/s

Step 2: Use the kinematic equation for vertical motion. The maximum

height (h) can be found using the vertical motion kinematic equation:

f=v2

i+ 2a·∆y

Where vf= 0 m/s (at the maximum height), vi=v0y,a=−9.81 m/s2(accel-

eration due to gravity), and ∆y=h. Substitute these values into the equation:

0 = (25.98)2+ 2(−9.81)h

Step 3: Solve for the maximum height. Solving the equation from Step 2 for

h, we get:

h=(25.98)2

2(9.81) ≈33.14 m

Therefore, the maximum height the projectile reaches during its flight is

approximately 33.14 meters.

Question 24

Question

A baseball pitcher throws a fastball from the mound towards home plate at

an angle of 20◦above the horizontal with an initial speed of 40 m/s. At that

instant, the batter is 18 meters away from the pitcher in the horizontal direction.

Calculate:

1. The time it takes for the ball to reach the batter.

2. The height of the ball when it reaches the batter.

Assume the ball is caught at the same height it was pitched.

Solution

1. Let’s first find the time it takes for the ball to reach the batter. We can

analyze the horizontal and vertical components of the motion separately.



Horizontal Motion: The initial velocity in the horizontal direction is

v0x= 40 cos(20◦) m/s and the horizontal displacement is ∆x= 18 m. We

can use the equation vx=∆x

tto find the time t.

Step 1: Calculate the horizontal component of initial velocity.

v0x= 40 cos(20◦)≈37.19 m/s

Step 2: Use the equation vx=∆x

tto find t.

t=∆x

v0x

=18

37.19 ≈0.48 s

Therefore, it takes approximately 0.48 seconds for the ball to reach the

batter.

2. Now, let’s find the height of the ball when it reaches the batter. We can

use the vertical motion to find the height at that point.



Vertical Motion: The initial velocity in the vertical direction is v0y=

40 sin(20◦) m/s. We can use the kinematic equation y=y0+v0yt−1

2gt2

to find the height ywhen t= 0.48 s.

Step 1: Calculate the vertical component of initial velocity.

v0y= 40 sin(20◦)≈13.68 m/s

Step 2: Use the kinematic equation y=y0+v0yt−1

2gt2to find the height

y. Since the ball is caught at the same height it was pitched, y0= 0 and

y= 0.

0 = 0 + 13.68 ×0.48 −1

2×9.81 ×(0.48)2

0=6.56 −1.13 ≈5.44 m

Therefore, the height of the ball when it reaches the batter is approximately

5.44 meters.

Question 25

Question

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

at an angle of 30◦above the horizontal. Determine the maximum height the

projectile reaches during its flight.

Solution

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

The horizontal component is given by

vix=vicos(θ)

where vixis the initial horizontal velocity, viis the initial velocity (given as

30 m/s), and θis the angle of launch (given as 30◦). Substituting the values:

vix= 30 m/s ·cos(30◦)≈25.98 m/s

Step 2: The vertical component of the initial velocity can be determined

using

viy=visin(θ)

where viyis the initial vertical velocity. Substituting the values:

viy= 30 m/s ·sin(30◦)≈15 m/s

Step 3: Use the vertical motion equation to find the time taken to reach

maximum height. At the highest point, the vertical velocity is zero. So, using

the equation

vf=vi+a·t

where vf= 0 m/s (at the top of the motion), vi= 15 m/s, and a=−9.8 m/s2,

solve for t:

0 = 15 m/s −9.8 m/s2·t

t=15 m/s

9.8 m/s2≈1.53 s

Step 4: Substitute the time into the vertical position equation to find the

maximum height. The vertical position equation is given by

y=yi+viy·t+1

2a·t2

where yi= 0, viy= 15 m/s, a=−9.8 m/s2, and t= 1.53 s. Solving for y:

y= 0 + 15 m/s ·1.53 s + 1

2(−9.8 m/s2)·(1.53 s)2≈11.0 m

Therefore, the maximum height the projectile reaches during its flight is

approximately 11.0 m.

Question 26

Question

A baseball is hit with an initial speed of 30 m/s at an angle of 45 degrees above

the horizontal. How far away does it land? Ignore air resistance and assume

the baseball is hit from ground level.

Solution

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

nents. The initial velocity of the baseball can be written as: Vix=Vicos θ=

(30 m/s) cos(45◦)Viy=Visin θ= (30 m/s) sin(45◦)

Step 2: Find the time it takes for the baseball to hit the ground. The

time it takes for the baseball to hit the ground can be found using the equation:

y=Viyt+1

2ayt2Where y= 0 (final position), Viy= 30 m/s·sin(45◦), ay=−9.81

m/s2. Solve for t.

Step 3: Find the horizontal distance the baseball travels. The horizontal dis-

tance the baseball travels can be found using the equation: x=VixtSubstitute

the value of t found in step 2 and solve for x.

Question 27

Question

A particle moves in a plane with a constant acceleration. At t= 0, its position

vector is r(0) = 2ˆ

i−3ˆ

jm and its velocity vector is v(0) = 4ˆ

i+ 2ˆ

jm/s. If its

acceleration vector is a= 2ˆ

i+ 3ˆ

jm/s2, find its position vector at t= 3 s.

Solution

Step 1: First, we need to find the particle’s velocity vector at time t= 3 s using

the formula v(t) = v(0) + at.

Given: v(0) = 4ˆ

i+ 2ˆ

jm/s, a= 2ˆ

i+ 3ˆ

jm/s2,t= 3 s.

Substitute the values into the formula:

v(3) = 4ˆ

i+ 2ˆ

j+ (2ˆ

i+ 3ˆ

j)∗3

= 4ˆ

i+ 2ˆ

j+ 6ˆ

i+ 9ˆ

= 10ˆ

i+ 11ˆ

Therefore, v(3) = 10ˆ

i+ 11ˆ

jm/s.

Step 2: Next, we can find the particle’s position vector at time t= 3 s using

the formula r(t) = r(0) + v(0)t+1

2at2.

Given: r(0) = 2ˆ

i−3ˆ

jm, v(0) = 4ˆ

i+ 2ˆ

jm/s, a= 2ˆ

i+ 3ˆ

jm/s2,t= 3 s.

Substitute the values into the formula:

r(3) = (2ˆ

i−3ˆ

j) + (4ˆ

i+ 2ˆ

j)∗3 + 1

2(2ˆ

i+ 3ˆ

j)∗32

= 2ˆ

i−3ˆ

j+ 12ˆ

i+ 6ˆ

j+1

2(6ˆ

i+ 9ˆ

= 2ˆ

i−3ˆ

j+ 12ˆ

i+ 6ˆ

j+ 3ˆ

i+9

2ˆ

= 17ˆ

i+9

2ˆ

Therefore, r(3) = 17ˆ

i+9

2ˆ

jm.

Question 28

Question

A projectile is launched from the ground at an angle of 45◦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 falls vertically with zero initial

speed while the other fragment moves in the opposite direction of the original

projectile with the same speed as the original projectile. Find the distance

between the fragments just before the explosion. (Neglect air resistance, assume

g= 9.8 m/s2)

Solution

Step 1: Find the time of flight of the original projectile. Let xand ybe the

horizontal and vertical components of the original projectile’s displacement,

respectively. The time of flight Tcan be found using the equation for the

vertical displacement:

y=visin(θ)T−1

2gT 2

Substitute vi= 20 m/s, θ= 45◦,y= 0, and g= 9.8 m/s2:

0 = 20 sin(45◦)T−1

2·9.8·T2

T=40

9.8s≈4.08 s

Step 2: Find the horizontal distance the original projectile travels. The

horizontal distance Xthe original projectile travels can be found using the

equation for the horizontal displacement:

X=vicos(θ)T

Substitute vi= 20 m/s, θ= 45◦, and T≈4.08 s:

X= 20 cos(45◦)·4.08 ≈57.77 m

Step 3: Find the distance between the fragments just before the explosion.

At the highest point, the vertical component of the velocity of the original pro-

jectile is zero. This means the two fragments of equal mass separate horizontally

to overcome their initial speed 20 m/s, until one fragment falls vertically. The

horizontal distance each fragment traveled is half of the horizontal distance of

the original projectile. Thus, the distance between the fragments just before

the explosion is X

2=57.77

2= 28.88 m.

Question 29

Question

A baseball pitcher throws a ball at an angle of 30◦above the horizontal. The

ball leaves the pitcher’s hand at a speed of 30 m/s.

1. Find the maximum height above the pitcher’s hand that the ball reaches.

2. Find the total time the ball is in the air.

Solution

1. To find the maximum height above the pitcher’s hand, we can analyze the

vertical motion of the ball. The key point to note is that at the maximum

height, the vertical component of the ball’s velocity is zero. We will use the

kinematic equation for vertical motion:

vf=vi+at

where: vf= 0 (final vertical velocity at maximum height), vi=viy=vsin θ

(initial vertical velocity), a=−g(acceleration due to gravity is negative as it

acts downward), and we want to solve for t.

Step 1: Substitute the known values into the kinematic equation:

0 = 30 sin(30◦)−9.8t

Step 2: Solve for t:

9.8t= 30 sin(30◦)

t=30 sin(30◦)

9.8

t≈1.53 seconds

Step 3: To find the maximum height, we can use the vertical position

equation:

y=y0+viyt+1

2at2

where: y= 0 (final vertical position at maximum height), y0= 0 (initial vertical

position), viy=vsin θ(initial vertical velocity), a=−g(acceleration due to

gravity), and we want to solve for y.

Step 4: Substitute the known values into the vertical position equation:

0 = 0 + 30 sin(30◦)×1.53 −1

2×9.8×(1.53)2

Step 5: Solve for y:

y≈11.3 meters

Therefore, the maximum height above the pitcher’s hand that the ball reaches

is approximately 11.3 meters.

2. The total time the ball is in the air consists of the time to reach the max-

imum height and the time to come back down. Since the motion is symmetric,

the total time is twice the time to reach the maximum height.

Step 6: Calculate the total time the ball is in the air:

Total time = 2 ×1.53

Total time = 3.06 seconds

Thus, the total time that the ball is in the air is 3.06 seconds.

Question 30

Question

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

speed of 20 m/s. Determine the total time that the projectile is in the air.

Neglect air resistance and assume g= 9.81 m/s2.

Solution

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

The initial velocity of the projectile is given by v0= 20 m/s at an angle of

θ= 30◦above the horizontal. The horizontal component is v0x=v0cos θ

and the vertical component is v0y=v0sin θ. Given: v0= 20 m/s θ= 30◦

g= 9.81 m/s2

We find that:

Step 2: Determine the time the projectile spends in the air. The time for

the projectile to reach the highest point can be found using the vertical motion

equation:

vy=v0y−gt

After reaching the highest point, the vertical component of the velocity becomes

zero. Thus, we have:

0 = 10 −9.81t

Solving for tgives:

t=10

9.81 ≈1.02 s

Step 3: Calculate the total time of flight. The total time of flight is twice

the time to reach the highest point because the time from the highest point to

the ground is equal to the time to reach the highest point:

Ttotal = 2t≈2(1.02) ≈2.04 s

Therefore, the total time that the projectile is in the air is approximately

2.04 seconds.

Question 31

Question

A soccer player kicks a soccer ball from the ground at an angle of 30◦above

the horizontal. The ball lands on the ground 60 m away from the player, after

being in the air for 3 seconds. Assuming air resistance is negligible, calculate:

1. The initial velocity of the soccer ball.

2. The maximum height the soccer ball reaches during its flight.

Solution

1. To find the initial velocity of the soccer ball, we can analyze the motion in

the horizontal and vertical directions separately.

Step 1: Break down the initial velocity Let v0xand v0ybe the hori-

zontal and vertical components of the initial velocity v0. Since the angle above

the horizontal is 30◦, we have:

v0x=v0cos 30◦

v0y=v0sin 30◦

Step 2: Determine the horizontal motion The horizontal distance the

ball travels can be calculated using the equation for horizontal motion:

d=v0xt

Substitute the given values d= 60 m and t= 3 s:

60 = v0cos 30◦×3

v0=60

3 cos 30◦

Step 3: Calculate the initial velocity Using the value of v0determined

above, we find:

v0=60

3 cos 30◦≈34.64 m/s

Therefore, the initial velocity of the soccer ball is approximately 34.64 m/s.

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

kinematic equation for vertical motion.

Step 4: Determine the vertical motion The vertical displacement of

the soccer ball is given by:

ymax =v0yt−1

2gt2

At the maximum height, the vertical component of the velocity is zero, so v0y=

v0sin 30◦.

Step 5: Calculate the maximum height Substitute the values v0=

34.64 m/s and t= 3 s, and g= 9.81 m/s2:

ymax = 34.64 sin 30◦×3−1

2×9.81 ×32

ymax = 52.5−44.145

ymax ≈8.355 m

Therefore, the maximum height the soccer ball reaches during its flight is

approximately 8.355 m.

Question 32

Question

A particle moves in a plane according to the equation r(t) = (cos t)ˆ

i+ (sin t)ˆ

where tis in seconds. Find the velocity and acceleration vectors of the particle

at t=π/4 seconds.

Solution

Step 1: To find the velocity vector, we differentiate the position vector r(t) with

respect to time t.

Step 1: v(t) = dr

Step 2: First differentiate the xand ycomponents of r(t) with respect to t.

Step 2: d(cos t)

dt =−sin tand d(sin t)

dt = cos t

Step 3: Combine the derivatives of the components to find the velocity

vector.

Step 3: v(t) = −sin tˆ

i+ cos tˆ

Step 4: Substitute t=π/4 into the velocity vector.

Step 4: v(π/4) = −√2

2ˆ

i+√2

2ˆ

Step 5: To find the acceleration vector, differentiate the velocity vector v(t)

with respect to time t.

Step 5: a(t) = dv

Step 6: Differentiate the xand ycomponents of v(t) with respect to t.

Step 6: d(−sin t)

dt =−cos tand d(cos t)

dt =−sin t

Step 7: Combine the derivatives of the components to find the acceleration

vector.

Step 7: a(t) = −cos tˆ

i−sin tˆ

Step 8: Substitute t=π/4 into the acceleration vector.

Step 8: a(π/4) = −√2

2ˆ

i−√2

2ˆ

Question 33

Question

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

of 50 m/s. How far does the ball travel horizontally before hitting the ground?

(Assume the ground is level.)

Solution

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

The initial velocity can be resolved into horizontal and vertical components

using trigonometry. The horizontal component is given by v0x=v0cos(θ) and

the vertical component is given by v0y=v0sin(θ), where v0= 50 m/s and

θ= 30◦.

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

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

gravity. The equation to use is vf y =v0y−gt, where vf y = 0 m/s (final vertical

velocity when the ball hits the ground), v0y= 50 sin(30◦) m/s, g= 9.81 m/s2,

and tis the time of flight.

Step 3: Calculate the horizontal distance. The horizontal distance traveled

by the ball can be found using the horizontal component of the initial velocity

and the time of flight. The equation to use is x=v0xt, where v0x= 50 cos(30◦)

m/s.

Step 4: Substitute the values and calculate the horizontal distance. Substi-

tute the calculated values into the equation for the horizontal distance to find

the distance traveled by the golf ball before hitting the ground.

Step 5: Calculate the final answer. Perform the final calculation to determine

how far the golf ball travels horizontally before hitting the ground.

Question 34

Question

A car travels along a path described by the curve y=x2, where xand yare in

meters. At a certain instant, the car’s position is given by x= 4 m and y= 16 m.

If the car’s velocity vector at this instant is vx= 3 m/s in the positive xdirection

and vy= 6 m/s in the positive ydirection, what is the car’s acceleration vector

at this instant?

Solution

Step 1: Find the unit tangent vector. The unit tangent vector to the path is

given by

T(t) = v(t)

∥v(t)∥

where v(t) is the velocity vector. Given that vx= 3 m/s and vy= 6 m/s, we

have

∥v(t)∥=qv2

x+v2

y=p32+ 62=√45 = 3√5 m/s

So, the unit tangent vector is

T=vx

∥v(t)∥,vy

∥v(t)∥=3

3√5,6

3√5=1

√5,2

√5

Step 2: Find the unit normal vector. The unit normal vector to the path is

given by

N(t) =

dt∥dT(t)

dt ∥Since the motion is along a curve y=x2, the derivative of T(t) with

respect to trepresents the curvature of the path. Therefore, dT(t)

dt is directed

toward the center of curvature of the path. Since the car travels along the path

y=x2, the unit normal vector is perpendicular to the path at all times and

points inward towards the concave side of the path.

Step 3: Find the acceleration vector. The acceleration vector can be split

into two components: centripetal acceleration and tangential acceleration. The

centripetal acceleration is given by v2

r, where vis the speed of the car and r

is the radius of curvature. The tangential acceleration is the rate of change of

the speed of the car. The total acceleration vector is the sum of the centripetal

acceleration and the tangential acceleration.

Since we know the unit tangent vector and the unit normal vector, the

acceleration vector can be written as:

a(t) = aTT+aNN

where aTis the tangential component of acceleration and aNis the normal

component of acceleration.

At this instant, the car’s acceleration vector is:

a=aTT+aNN

Question 35

Question

A baseball player is trying to steal second base. He starts from first base and

runs with a speed of 6.0 m/s. The second baseman has the ball and is standing

42 m from the player at an angle of 30 degrees with respect to the line connecting

the bases. If the player continues running at a constant speed, will he be safe

or out?

a(t)=6ˆ

Step 5: Now, we can find the acceleration of the particle at t= 2 s.

a(2) = 6ˆ

Step 6: To find the magnitude of the acceleration, we calculate |a(2)|.

|a(2)|=p(6)2= 6 m/s2

Therefore, at t= 2 s, the magnitude of the particle’s velocity is 4√10 m/s

and the magnitude of the acceleration is 6 m/s2.

Question 2

Question

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

speed of 50 m/s. Calculate the maximum height reached by the projectile.

Solution

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

The initial velocity can be broken into horizontal and vertical components as

follows: Vix=Vi·cos(30◦) = 50 m/s ·cos(30◦)Viy=Vi·sin(30◦) = 50 m/s ·

sin(30◦)

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

taken for the projectile to reach the maximum height can be determined using

the vertical component of the initial velocity and the acceleration due to gravity:

Vfy= 0 m/s (at maximum height) Using the kinematic equation Vfy=Viy−gt,

where g= 9.81 m/s2: 0 = 50 m/s ·sin(30◦)−9.81 m/s2·tSolving for tgives us

the time taken to reach the maximum height.

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

using the vertical motion equation: Hmax =Viy·t−1

2gt2Substitute the values

of Viyand tinto the equation to find the maximum height.

Question 3

Question

A soccer player kicks a ball from ground level with an initial velocity of 20 m/s

at an angle of 45◦above the horizontal. Calculate the maximum height the ball

reaches and the total time of flight.

Solution

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

The initial velocity vi= 20 m/s can be resolved into horizontal and vertical

components as follows: Horizontal component: vix=vicos(45◦) = 20 ·√2

10√2 m/s Vertical component: viy=visin(45◦) = 20 ·√2

2= 10√2 m/s

Step 2: Calculate the time it takes for the ball to reach the maximum height.

Using the vertical motion equation vf=vi+at where vf= 0 m/s (at the

maximum height) and a=−9.81 m/s2(acceleration due to gravity), we have:

0 = 10√2−9.81tSolving for t, we get: t=10√2

9.81 s

Step 3: Calculate the maximum height the ball reaches. Using the vertical

motion equation y=vit+1

2at2with vi= 10√2 m/s, a=−9.81 m/s2and t

from Step 2, we have: y= 10√2·10√2

9.81 +1

2(−9.81) 10√2

9.81 2Solving for y, we

get: y=(200)

9.81 −(200)

9.81 =200

9.81 m

Step 4: Calculate the total time of flight. Since the motion is symmetrical,

the time of flight is twice the time it takes to reach the maximum height: Total

time of flight = 2 ·10√2

9.81 s

Question 4

Question

A projectile is launched from the ground with an initial speed of 40 m/s at an

angle of 30 degrees above the horizontal. Find the maximum height above the

ground reached by the projectile. (Neglect air resistance and take g= 9.8 m/s2)

Solution

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

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

components as follows:

v0x=v0cos(θ) = 40 cos(30◦)≈34.64 m/s

v0y=v0sin(θ) = 40 sin(30◦)≈20 m/s

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

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

the equation of motion: vf=vi+at, where vf= 0, vi= 20 m/s, a=−9.8

m/s2, and tis the time taken to reach the maximum height, we have:

0 = 20 −9.8t

t=20

9.8≈2.04 s

Step 3: Calculate the maximum height reached by the projectile. Using the

equation of motion: s=vit+1

2at2, where sis the displacement in the vertical

direction, vi= 20 m/s, a=−9.8 m/s2, and t= 2.04 s, we have:

s= 20(2.04) + 1

2(−9.8)(2.04)2

s≈20.4 m

Therefore, the maximum height above the ground reached by the projectile

is approximately 20.4 meters.

Question 5

Question

A quarterback throws a football at an angle of 45◦above the horizontal with

an initial speed of 20 m/s. How far away does the football land from the

quarterback?

Solution

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

The initial velocity can be broken down into its horizontal component (vi,x)

and vertical component (vi,y ) using trigonometry. Since the angle is 45◦, both

components are equal.

vi,x =vi,y =vi·cos 45◦=vi·sin 45◦= 20 m/s ·1

√2≈14.14 m/s

Step 2: Determine the time of flight. The time taken for the football to

reach the highest point can be found using the vertical component of the initial

velocity and acceleration due to gravity.

vf,y =vi,y −g·t

0 = 14.14 m/s −9.81 m/s2·t

t=14.14 m/s

9.81 m/s2≈1.44 s

Step 3: Calculate the horizontal distance. The horizontal distance can be

calculated using the horizontal component of the initial velocity and the time

of flight.

Distance = vi,x ·t

Distance = 14.14 m/s ·1.44 s ≈20.36 m

Therefore, the football lands approximately 20.36 meters away from the

quarterback.

Question 6

Question

A baseball player hits a ball with an initial velocity of 40 m/s at an angle of

30◦above the horizontal. The ball lands on the ground 150 meters away. How

long is the ball in the air? What is the maximum height the ball reaches?

Solution

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

The initial velocity of the ball can be resolved into its horizontal (v0x) and

vertical (v0y) components using trigonometry:

v0x=v0cos(θ)

v0y=v0sin(θ)

Where v0= 40 m/s and θ= 30◦.

Substitute the values to find v0xand v0y:

v0x= 40 cos(30◦) = 34.64 m/s

v0y= 40 sin(30◦) = 20 m/s

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

the equation for time taken for an object to fall freely from an initial height to

the ground:

t=2v0y

Where g= 9.81 m/s2is the acceleration due to gravity.

Substitute v0yinto the equation to find t:

t=2×20

9.81 = 4.08 s

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

the ball can be determined using the equation for vertical motion:

h=v2

0y/(2g)

Substitute the values of v0yand gto calculate the maximum height:

h= (202)/(2 ×9.81) = 20.41 m

Therefore, the ball is in the air for 4.08 seconds and reaches a maximum

height of 20.41 meters.

Question 7

Question

A baseball is hit with an initial speed of 30 m/s at an angle of 30 degrees above

the horizontal. How far does the baseball travel horizontally before landing on

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

Solution

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

The initial velocity v0can be resolved into its horizontal component v0xand

vertical component v0yas follows:

v0x=v0cos θ

v0y=v0sin θ

where v0= 30 m/s and θ= 30◦. Plugging in the values, we get:

v0x= 30 m/s ×cos 30◦= 26 m/s

v0y= 30 m/s ×sin 30◦= 15 m/s

Step 2: Determine the time of flight (t). The time of flight for the baseball

can be found using the formula for vertical motion:

∆y=v0yt+1

2at2

where ∆yis the vertical displacement (equal to the height the baseball is hit

from, which we’ll assume is 0), a=−g(negative because it is acting in the

downward direction). Rearranging for t, we have:

0 = 15t−1

2×9.8×t2

Solving this quadratic equation gives t= 3.06 s.

Step 3: Calculate the horizontal distance traveled (∆x). The horizontal

distance ∆xcan be found using the formula for horizontal motion:

∆x=v0x×t

Plugging in the values, we get:

∆x= 26 m/s ×3.06 s = 79.56 m

Therefore, the baseball travels approximately 79.56 meters horizontally be-

fore landing on the ground.

Question 8

Question

A ball is thrown horizontally from the top of a building that is 50 meters high.

The initial velocity of the ball is 20 m/s. How far from the base of the building

does the ball land?

Solution

Step 1: Identify the known variables. The initial vertical velocity of the ball

is 0 m/s (thrown horizontally). The final vertical position of the ball is 50 m

(height of the building). The acceleration in the vertical direction is due to

gravity, which is −9.8 m/s2. The initial horizontal velocity of the ball is 20 m/s

and the ball lands on the ground at some horizontal distance we need to find.

Step 2: Determine the time taken for the ball to reach the ground. Since

the ball is only influenced by gravity in the vertical direction, we can use the

equation:

yf=yi+vyit+1

2at2.

Plug in the known values:

50 m = 0 m + 0 m/s ·t+1

2·(−9.8 m/s2)·t2.

Solving for t, we get:

t=s2·50 m

9.8 m/s2≈3.20 s.

Step 3: Calculate the horizontal distance traveled by the ball. Since the

horizontal velocity of the ball is constant, we can use the equation:

xf=xi+vxit.

Plug in the known values:

xf= 0 m + 20 m/s ·3.20 s = 64 m.

Answer: The ball lands 64 meters from the base of the building.

Question 9

Question

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

The initial speed of the cannonball is 50 m/s. Calculate the following:

1. The maximum height the cannonball reaches.

2. The time it takes for the cannonball to reach the maximum height.

3. The total time the cannonball is in the air before hitting the ground.

(Assume no air resistance.)

Solution

Let’s denote the initial speed of the cannonball as v0= 50 m/s, the launch angle

as θ= 30◦, the acceleration due to gravity as g= 9.81 m/s2, and the maximum

height reached by the cannonball as H. We will use the kinematic equations to

solve for the different parts of the question.

Part 1: Maximum height reached by the cannonball

Step 1: We can determine the vertical component of the initial velocity

(v0y) using the initial speed and launch angle:

v0y=v0sin(θ)

v0y= 50 m/s ·sin(30◦)

v0y≈25 m/s

Step 2: Next, we can use the kinematic equation for vertical motion to find

the maximum height:

fy=v2

0y−2·g·H

At the maximum height, vfy= 0.

0 = (25 m/s)2−2·9.81 m/s2·H

H=(25 m/s)2

2·9.81 m/s2

H≈31.93 m

Thus, the maximum height reached by the cannonball is approximately 31.93

meters.

Part 2: Time to reach the maximum height

Step 1: We can find the time it takes for the cannonball to reach the

maximum height using the vertical component of the initial velocity:

vfy=v0y−g·t

At the maximum height, vfy= 0.

0 = 25 m/s −9.81 m/s2·t

t=25 m/s

9.81 m/s2

t≈2.55 s

Therefore, it takes approximately 2.55 seconds for the cannonball to reach

the maximum height.

Part 3: Total time in the air

Since the motion is symmetric, the total time the cannonball is in the air

before hitting the ground is twice the time it takes to reach the maximum height.

Total time = 2 ×(Time to reach maximum height)

Total time = 2 ×2.55 s

Total time ≈5.10 s

Therefore, the total time the cannonball is in the air before hitting the

ground is approximately 5.10 seconds.

Question 10

Question

A baseball is hit with an initial velocity of 30 m/s in the horizontal direction

and 20 m/s in the vertical direction. Calculate the baseball’s maximum height,

the time it takes to reach the highest point, and the total time of flight. Assume

air resistance is negligible and take the acceleration due to gravity as 9.8 m/s2.

Solution

Step 1: Find the time to reach the highest point. Given that the initial ver-

tical velocity of the baseball is 20 m/s and the acceleration due to gravity is

−9.8 m/s2, we can use the kinematic equation:

vf=vi+at

where vfis the final velocity (0 m/s at the highest point), viis the initial velocity

(20 m/s up), ais the acceleration due to gravity (-9.8 m/s2), and tis the time

taken to reach the highest point. Substitute the known values:

0 = 20 −9.8t

Solving for t:

t=20

9.8≈2.04 s

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

for vertical motion:

yf=yi+vit+1

2at2

where yfis the final height (maximum height), yiis the initial height (0 m), vi

is the initial vertical velocity (20 m/s), ais the acceleration due to gravity (-9.8

m/s2), and tis the time taken to reach the highest point (2.04 s). Substitute

the known values:

yf= 0 + 20 ×2.04 −1

2×9.8×(2.04)2

yf≈20 ×2.04 −1

2×9.8×4.1616

yf≈40.8−20.4096

yf≈20.3904 m

Step 3: Determine the total time of flight. The total time of flight is twice

the time taken to reach the highest point.

T otal time = 2 ×2.04

T otal time = 4.08 s

Therefore, the baseball’s maximum height is 20.39 m, the time it takes to

reach the highest point is approximately 2.04 s, and the total time of flight is

4.08 s.

Question 11

Question

A soccer player kicks a ball from the ground with an initial speed of 20 m/s at

an angle of 30 degrees above the horizontal. The ball lands on the ground 4

seconds later. (a) What is the horizontal distance the ball travels? (b) What is

the maximum height the ball reaches? (c) What are the horizontal and vertical

components of the ball’s velocity when it lands?

Solution

Step 1: Determine the horizontal distance the ball travels.

Given: v0= 20 m/s, θ= 30◦,t= 4 s

The horizontal component of the initial velocity is v0x=v0cos(θ) and the

vertical component of the initial velocity is v0y=v0sin(θ).

Using the horizontal distance formula: x=v0x·t, we can find x.

v0x= 20 m/s ·cos(30◦) = 20 m/s ·√3

2= 10√3 m/s

x= 10√3 m/s ·4 s = 40√3 m ≈69.3 m

Therefore, the horizontal distance the ball travels is approximately 69.3 me-

ters.

Step 2: Determine the maximum height the ball reaches.

Using the vertical distance formula: y=v0yt−1

2gt2, where g= 9.8 m/s2.

Substitute the given values into the formula to find the maximum height,

ymax.

v0y= 20 m/s ·sin(30◦) = 20 m/s ·1

2= 10 m/s

ymax = 10 m/s ·4 s −1

2·9.8 m/s2·(4 s)2

ymax = 40 m −78.4 m = −38.4 m

The negative sign indicates the ball is below the initial point, so the max-

imum height the ball reaches is approximately 38.4 meters below the initial

point.

Step 3: Find the horizontal and vertical components of the ball’s velocity

when it lands.

The vertical component of the velocity when the ball lands is the same as the

initial vertical component due to symmetry. Therefore, the vertical component

is vfinaly=−10 m/s.

Since there is no horizontal acceleration, the horizontal component of the

velocity remains constant. Therefore, the horizontal component is vfinalx=

10√3 m/s.

Therefore, the horizontal and vertical components of the ball’s velocity when

it lands are 10√3 m/s and −10 m/s respectively.

Question 12

Question

A projectile is launched with an initial velocity of 30 m/s at an angle of 45◦

above the horizontal. Determine the maximum height the projectile reaches.

Given: g= 9.81 m/s2

Solution

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

The initial velocity, vi, can be resolved into horizontal and vertical components

using trigonometric functions. The vertical component is:

viy=visin θ

viy= 30 sin 45◦

viy≈21.21 m/s

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

to reach maximum height (tmax) can be found using the vertical component of

the initial velocity and the acceleration due to gravity. At maximum height, the

vertical component of velocity becomes 0 m/s.

vfy= 0 m/s = viy−g·tmax

0 = 21.21 −9.81 ·tmax

tmax ≈2.16 s

Step 3: Calculate the maximum height. The maximum height (Hmax) can

be determined using the kinematic equation for vertical motion:

Hmax =viy·tmax −1

2g·t2

max

Hmax = 21.21 ·2.16 −1

2·9.81 ·(2.16)2

Hmax ≈23.10 m

Therefore, the maximum height the projectile reaches is approximately 23.10

meters.

Question 13

Question

A projectile is launched from ground level at an angle of 60◦above the horizontal

with an initial speed of 20 m/s.

(a) Determine the maximum height reached by the projectile.

(b) Find the total time of flight.

Solution

(a) To find the maximum height reached by the projectile, we need to determine

the vertical component of the initial velocity.

Step 1: Find the initial vertical velocity component (viy). Given that the

initial speed is 20 m/s and the launch angle is 60◦, we have:

viy=visin(θ) = 20 m/s ×sin(60◦).

Therefore, viy= 20 ×√3

2≈17.32 m/s.

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

height, the vertical velocity component is 0 m/s. Using the equation vf=vi+at,

we have:

0 = 17.32 −9.81t.

Solving for tgives us:

t=17.32

9.81 ≈1.77 s.

Step 3: Find the maximum height. The maximum height (hmax) can be

determined using the equation:

hmax =viyt−1

2gt2.

Substitute the known values to get:

hmax = 17.32 ×1.77 −1

2×9.81 ×(1.77)2.

Calculating this gives hmax ≈15.15 m.

Therefore, the maximum height reached by the projectile is approximately

15.15 meters.

(b) To find the total time of flight, we can use the fact that the time to reach

maximum height is half the total time of flight.

The total time of flight is:

Total time = 2 ×1.77 ≈3.54 s.

formula:

Range = vix×Total time of flight,

where vixis the initial horizontal velocity component. This can be calculated

as:

vix=vicos(θ) = 20 m/s ×cos(60◦).

Thus, vix= 20 ×1

2= 10 m/s.

Substitute the values to get:

Range = 10 ×3.54 = 35.4 m.

Therefore, the horizontal range of the projectile is 35.4 meters.

Question 14

Question

An object is launched from ground level with an initial velocity of 20 m/s at an

angle of 30◦above the horizontal. Find the maximum height reached by the

object.

Solution

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

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

components as follows:

Vix=Vicos θ= 20 m/s cos 30◦

Viy=Visin θ= 20 m/s sin 30◦

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

to reach maximum height can be found using the equation v=vi+at. In

the vertical direction, the final velocity at maximum height is 0 m/s, the initial

velocity is Viy, the acceleration is due to gravity (−9.8 m/s2), and the time is

unknown.

0 = Viy−9.8 m/s2·t

Solving for t, we get:

t=Viy

9.8 m/s2

Step 3: Find the maximum height reached by the object. The maximum

height reached can be calculated using the equation h=Viyt−1

2gt2, where his

the maximum height.

h=Viy·Viy

9.8 m/s2−1

2·(−9.8 m/s2)· Viy

9.8 m/s2!2

Simplify the expression to find the maximum height reached by the object.

Question 15

Question

A projectile is launched at an angle of 45◦above the horizontal with an initial

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

Solution

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

Let v0= 20 m/s be the initial speed. The horizontal component of the initial

velocity v0x=v0cos(45◦) and the vertical component of the initial velocity

v0y=v0sin(45◦).

Step 2: Calculate the time to reach the maximum height. Consider the ver-

tical motion of the projectile. At the maximum height, the vertical component

of the velocity will be zero. Use the equation vyf =v0y−gt where vyf = 0 m/s.

Solve for tto find the time to reach the maximum height.

Step 3: Calculate the maximum height. Using the vertical motion equation

y=y0+v0yt−1

2gt2, where y0= 0 m (starting from the ground) and the time

tcalculated in Step 2, find the maximum height yreached by the projectile.

Step 4: Substitute the known values and solve for the maximum height.

Substitute v0y,t, and g= 9.81 m/s2into the equation from Step 3 to find the

maximum height y.

Step 1:

The horizontal component of the initial velocity is given by:

v0x=v0cos(45◦) = 20 ×1

√2= 10√2 m/s

The vertical component of the initial velocity is given by:

v0y=v0sin(45◦) = 20 ×1

√2= 10√2 m/s

Step 2:

Using the equation vyf =v0y−gt and rearranging for t, we get:

0 = 10√2−9.81t

t=10√2

9.81 ≈1.43 s

Step 3:

Substitute y0= 0 m, v0y= 10√2 m/s, t= 1.43 s, and g= 9.81 m/s2into the

vertical motion equation:

y= 0 + 10√2×1.43 −1

2×9.81 ×(1.43)2

Step 4:

Calculating the maximum height:

y≈10.2 m

Therefore, the maximum height reached by the projectile is approximately

10.2 meters.

Question 16

Question

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

with an initial speed of 20 m/s. What are the maximum height above the ground

and the range of the projectile?

Solution

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

The initial velocity v0= 20 m/s can be resolved into its horizontal and vertical

components as: Horizontal component: v0x=v0cos 45◦Vertical component:

v0y=v0sin 45◦

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

using the vertical motion equation: y=v0yt−1

2gt2where yis the height,

g= 9.81 m/s2is the acceleration due to gravity, and the projectile reaches its

maximum height when vy= 0. Solving for time t, we get: 0 = v0yt−1

2gt2

t=2v0y

Step 3: Calculate the maximum height reached by the projectile. The max-

imum height hreached by the projectile can be calculated using the vertical

motion equation: h=v0yt−1

2gt2

Step 4: Calculate the range of the projectile. The range Rof the projectile

can be calculated using the horizontal motion equation: R=v0x×2t

Step 5: Substitute known values and calculate. Now, substitute the given

values (v0= 20 m/s, g= 9.81 m/s2,v0y= 20 ×sin 45◦,v0x= 20 ×cos 45◦) into

the formulas found in Steps 2, 3, and 4 to calculate the maximum height and

range of the projectile.

Question 17

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 of its trajectory, the projectile

explodes into two fragments of equal mass. One fragment, whose initial speed

is negligible, falls vertically from the highest point. Neglecting air resistance,

what is the speed of the other fragment just after the explosion?

Solution

Step 1: Determine the components of the initial velocity of the projectile. Given

that the initial speed of the projectile is 40 m/s and it is fired at an angle of 30◦

above the horizontal, we can find the initial xand ycomponents of the velocity

as follows:

v0x=v0cos(θ) = 40 cos(30◦) = 40 ·√3

2= 20√3 m/s

v0y=v0sin(θ) = 40 sin(30◦) = 40 ·1

2= 20 m/s

Step 2: Determine the time taken for the projectile to reach the highest

point. Since the vertical component of the initial velocity is zero at the highest

point, we can use the equation:

vy=v0y−gt

where vy= 0 and g= 9.81 m/s2(acceleration due to gravity). Solving for t, we

get:

0 = 20 −9.81t=⇒t=20

9.81 ≈2.04 s

Step 3: Find the maximum height reached by the projectile. Using the

formula for vertical displacement with constant acceleration:

y=v0yt−1

2gt2

Plugging in the values, we get:

y= 20 ·2.04 −1

2·9.81 ·(2.04)2= 40.8−1

2·9.81 ·4.1616 ≈40.8−20.48 ≈20.32 m

Step 4: Determine the velocity of the fragment falling vertically. Since the

other fragment falls vertically from the highest point, its velocity just after the

explosion is equal to the velocity of the ball at the highest point. Thus, its

velocity is given by the vertical component of the initial velocity:

v′

0y= 20 m/s

Step 5: Calculate the speed of the other fragment just after the explosion.

The speed of a vector is given by the magnitude of the vector. Therefore, the

speed of the other fragment just after the explosion is:

v′=q(v0x)2+ (v′

0y)2=q(20√3)2+ (20)2=√1200 + 400 = √1600 = 40 m/s

Therefore, the speed of the other fragment just after the explosion is 40 m/s.

Question 18

Question

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

with an initial speed of 40 m/s. Determine the maximum height the projectile

reaches during its flight.

Solution

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

The vertical component can be found using viy=visin θand the horizontal

component can be found using vix=vicos θ.

Given: Initial speed, vi= 40 m/s Angle, θ= 30◦

Therefore: viy= 40 m/s ×sin(30◦)≈20 m/s vix= 40 m/s ×cos(30◦)≈

34.64 m/s

Step 2: Calculate the time when the projectile reaches its maximum height

using the equation vf=vi+at for vertical motion, where the final vertical

velocity will be zero at the peak.

Given: Final vertical velocity, vfy= 0 Initial vertical velocity, viy= 20 m/s

Vertical acceleration, ay=−9.8 m/s2

Substitute the values into the equation and solve for t: 0 = 20 m/s −

9.8 m/s2×t

t=20 m/s

9.8 m/s2≈2.04 s

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

motion and the time calculated in Step 2.

Use the equation y=vit+1

2at2with vertical displacement y=?, initial

vertical velocity viy= 20 m/s, vertical acceleration ay=−9.8 m/s2, and time

t= 2.04 s.

Substitute the values into the equation: y= 20 m/s×2.04 s+ 1

2(−9.8 m/s2)(2.04 s)2

y≈20.40 m + (−20.00 m) ≈0.40 m

Therefore, the projectile reaches a maximum height of approximately 0.40

meters during its flight.

Question 19

Question

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

with an initial speed of 30 m/s. Find the maximum height reached by the

projectile during its flight.

Solution

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

The vertical component is given by:

V0y=V0·sin θ

V0y= 30 m/s ·sin 45◦≈21.21 m/s

Step 2: Use the kinematic equation for vertical motion to find the time it

takes for the projectile to reach its maximum height. At maximum height, the

vertical velocity is 0 m/s.

vy=v0y−gt

0 = 21.21 m/s −9.8 m/s2·t

t≈2.17 s

Step 3: Use the kinematic equation for vertical motion to find the maximum

height reached by the projectile.

y=v0yt−1

2gt2

y= 21.21 m/s ·2.17 s −1

2·9.8 m/s2·(2.17 s)2

y≈23.57 m

Therefore, the maximum height reached by the projectile during its flight is

approximately 23.57 meters.

Question 20

Question

A projectile is fired from the ground with an initial speed of 80 m/s at an angle

of 30 degrees 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 (vix) and vertical (viy)

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

components as follows:

vix=vi·cos(θ)

viy=vi·sin(θ)

where vi= 80 m/s and θ= 30◦. Thus,

vix= 80 ·cos(30◦) = 80 ·√3

2= 40√3 m/s

viy= 80 ·sin(30◦) = 80 ·1

2= 40 m/s

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

imum height, the vertical component of the projectile’s velocity becomes zero.

Using the vertical motion equation vf=vi+at, where vf= 0 and a=−9.8 m/s2

(acceleration due to gravity), we can solve for time t:

0 = 40 −9.8t

t=40

9.8≈4.08 s

Therefore, it takes approximately 4.08 seconds for the projectile to reach its

maximum height.