PHYS 202 - GENERAL PHYSICS II Kinematics in Two Dimensions Question Bank Set 5

PHYS 202 - GENERAL PHYSICS II -

Kinematics in Two Dimensions

Question Bank - Set 5

Liberty University

Question 1

Question

A projectile is fired with an initial speed of 100 m/s at an angle of 30◦above

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

Solution

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

The initial velocity components are: V0x=V0cos(θ) = 100 m/s ·cos(30◦)V0y=

V0sin(θ) = 100 m/s ·sin(30◦)

Calculating these values: V0x= 100 m/s·√3

2= 50√3 m/s V0y= 100 m/s·1

50 m/s

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

time to reach maximum height can be found using the vertical component:

Vfy =V0y−gt At maximum height, the vertical component of the velocity is 0,

so: 0 = 50 m/s −9.8 m/s2·t

Solving for t:t=50 m/s

9.8 m/s2≈5.10 s

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

be determined using the equation for vertical displacement: y=V0yt−1

2gt2

Substitute the known values: y= 50 m/s ·5.10 s −1

2·9.8 m/s2·(5.10 s)2

Evaluating this expression: y= 255 m −125.46 m = 129.54 m

Therefore, the maximum height reached by the projectile is approximately

129.54 m.

Question 2

Question

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

speed of 20 m/s. Determine the projectile’s maximum height above its launch

point.

Solution

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

The initial velocity v0can be resolved into its horizontal and vertical components

as:

v0x=v0cos(θ) = 20 m/s ·cos(30◦) = 17.32 m/s

v0y=v0sin(θ) = 20 m/s ·sin(30◦) = 10 m/s

Step 2: Use the vertical component of velocity to find the time to reach

maximum height. The time to reach maximum height can be found using the

vertical component of velocity:

vy=v0y−gt

where g= 9.8 m/s2is the acceleration due to gravity. At maximum height, the

vertical component of velocity is zero, so:

0 = 10 m/s −9.8 m/s2·t

Solving for t, we get:

t=10 m/s

9.8 m/s2= 1.02 s

Step 3: Calculate the maximum height. The maximum height (ymax) above

the launch point can be found using the kinematic equation:

ymax =v0yt−1

2gt2

Substitute the known values:

ymax = 10 m/s ·1.02 s −1

2·9.8 m/s2·(1.02 s)2

ymax = 10.20 m −5.10 m = 5.10 m

Therefore, the projectile’s maximum height above its launch point is 5.10

meters.

Question 3

Question

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

and lands 4 seconds later. If air resistance is neglected, determine the initial

speed of the projectile.

Solution

Step 1: Identify the known values and equations related to projectile motion.

In projectile motion, we can break the initial velocity into its horizontal (v0x)

and vertical (v0y) components. The key equations for projectile motion in two

dimensions are: 1. vx=v0x2. vy=v0y−gt 3. x=v0xt4. y=v0yt−1

2gt25.

y= (v0y−gt)2=v2

0y−2gy

Step 2: Determine the initial velocity components using trigonometric rela-

tionships.

Given that the projectile is launched at an angle of 30◦above the horizontal,

we can express the initial velocity as: v0x=v0cos(30◦)v0y=v0sin(30◦)

Step 3: Use the equation for time of flight to find the initial velocity.

The time of flight of the projectile is given as 4 seconds. This means the

projectile spends 4 seconds in the air, so t= 4s. Using the equation for

time: y=v0yt−1

2gt20 = v0yt−1

2gt2v0y=1

2gt Substitute v0y=1

2gt into

v0x=v0cos(30◦) and v0y=v0sin(30◦). vx=v0cos(30◦)vy=1

2g(4s)

Step 4: Calculate the initial speed of the projectile.

To find the initial speed, we need to find v0by summing the squares of the veloci-

ties in the x and y components: v0=qv2

x+v2

yv0=q(v0cos(30◦))2+1

2g(4s)2

Solve for v0.

Question 4

Question

A particle moves in the xy plane. Its position vector is given by r = (3t2−

2t)ˆ

i+ (4t+ 1)ˆ

j, where tis in seconds, xand yare in meters. Find the velocity

and acceleration vectors of the particle.

Solution

Step 1: To find the velocity vector v, we differentiate the position vector r with

respect to time t.

v =dr

Step 2: Differentiating each component of r with respect to t:

dt =d(3t2−2t)

dt = 6t−2

dt =d(4t+ 1)

dt = 4

Step 3: Therefore, the velocity vector v is:

v = (6t−2)ˆ

i+ 4ˆ

Step 4: To find the acceleration vector a, we differentiate the velocity vector

v with respect to time t.

a =dv

Step 5: Differentiating each component of v with respect to t:

d2x

dt2=d(6t−2)

dt = 6

d2y

dt2=d(4)

dt = 0

Step 6: Therefore, the acceleration vector a is:

a = 6ˆ

Question 5

Question

A particle moves in the xy plane with an acceleration given by a= (4t2ˆ

i−

6ˆ

j) m/s2, where tis in seconds. The particle’s velocity at t= 2.0 s is v=

(5.0ˆ

i−5.0ˆ

j) m/s. Find the particle’s position vector at t= 2.0 s if the particle

is at the origin when t= 0 s.

Solution

Step 1: To find the particle’s position vector at t= 2.0 s, we first need to find

the particle’s velocity and position functions in terms of time.

Given the acceleration a= (4t2ˆ

i−6ˆ

j) m/s2, we can find the velocity function

by integrating the acceleration function with respect to time:

v=Zadt =Z4t2dtˆ

i−Z6dtˆ

v=4

3t3+C1ˆ

i−6t+C2ˆ

Given that the velocity at t= 2.0 s is v= (5.0ˆ

i−5.0ˆ

j) m/s, plug in t= 2.0 s

to find the constants C1and C2:

(5.0ˆ

i−5.0ˆ

j) = 4

3(2.0)3+C1ˆ

i−6(2.0) + C2ˆ

(5.0ˆ

i−5.0ˆ

j) = (10.67 + C1)ˆ

i−12 + C2ˆ

Therefore, C1=−0.67 m/s in the ˆ

idirection and C2= 7.0 m in the ˆ

jdirection.

The velocity function is then:

v=4

3t3−0.67ˆ

i−6t+ 7ˆ

Step 2: To find the position function, integrate the velocity function with

respect to time:

r=Zvdt =Z4

3t3−0.67dtˆ

i−Z6t+ 7 dtˆ

r=1

3t4−0.67t+C3ˆ

i−3t2+ 7t+C4ˆ

Since the particle is at the origin when t= 0 s, we know that C3= 0 and

C4= 0.

Thus, the position function is:

r=1

3t4−0.67tˆ

i−3t2+ 7tˆ

Finally, plug in t= 2.0 s to find the position vector at t= 2.0 s:

r(2.0) = 1

3(2.0)4−0.67(2.0)

Question 6

Question

A cannonball is fired with an initial velocity of 200 m/s at an angle of 30◦above

the horizontal. Find the maximum height it reaches and the time taken to reach

that height.

Solution

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

The horizontal component of velocity, Vix, is given by: Vix=Vi·cos(θ) =

200 ·cos(30◦) = 173.21 m/s.

The vertical component of velocity, Viy, is given by: Viy=Vi·sin(θ) =

200 ·sin(30◦) = 100 m/s.

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

vertical motion equation vf=vi+at, where vf= 0 at the maximum height

and a=−9.81 m/s2: 0 = 100 −9.81t

t=100

9.81 = 10.20 s.

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

vertical motion equation h=vit+1

2at2:h= 100 ·10.20 + 1

2(−9.81)(10.20)2

h= 1020 −499.59 = 520.41 m.

Therefore, the maximum height reached by the cannonball is 520.41 m and

the time taken to reach that height is 10.20 s.

Question 7

Question

A baseball player hits a ball with an initial velocity of 30 m/s at an angle of 45◦

above the horizontal. The outfield wall is 100 m away and 5 m high. Will the

ball clear the wall? Assume there is no air resistance.

Solution

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

initial horizontal velocity v0xis given by:

v0x=v0cos θ= 30 m/s ×cos 45◦= 21.2 m/s

The initial vertical velocity v0yis given by:

v0y=v0sin θ= 30 m/s ×sin 45◦= 21.2 m/s

Step 2: Calculate the time of flight. Use the equation:

time = 2v0y

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

time = 2×21.2

9.81 = 4.33 s

Step 3: Calculate the maximum height reached by the ball. Use the equation:

hmax =v2

0y/2g

hmax = (21.2)2/(2 ×9.81) = 22.8 m

Step 4: Calculate the horizontal distance the ball travels. Using the equation:

d=v0x×time

d= 21.2×4.33 = 91.9 m

Step 5: Check if the ball clears the wall. Since the wall is 5 m high and the

ball reaches a maximum height of 22.8 m, the ball easily clears the 5 m wall.

Therefore, the ball will clear the 5 m high wall and reach a maximum height

of 22.8 m.

Question 8

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 what height above the point of projection

is the projectile 2 seconds after being launched?

Solution

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

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

components as follows: The initial horizontal component is vix=vicos(θ) where

viis the initial speed of the projectile and θis the launch angle. The initial

vertical component is viy=visin(θ).

Step 2: Determine the time taken for the projectile to reach a height of 2

m. The equation that relates the vertical displacement, initial vertical velocity,

time, and acceleration is:

y=viyt−1

2gt2

Since the projectile is at a height of 2 m, y= 2 m. Substitute viy= 30 sin(60◦) =

15√3 m/s and g= 9.8 m/s2into the equation.

2 = 15√3t−1

2·9.8·t2

0 = 15√3t−4.9t2

Solving for t, we get t= 0 (extraneous) or t≈0.349 s.

Step 3: Calculate the height above the point of projection after 2 seconds.

The equation for vertical position at time tcan be given by:

y=viyt−1

2gt2

At t= 2 s, substitute viy= 15√3 m/s, t= 2 s, and g= 9.8 m/s2into the

equation.

y= 15√3·2−1

2·9.8·22

y= 30√3−19.6

y≈37.38 m

Therefore, the projectile is approximately 37.38 m above the point of pro-

jection after 2 seconds.

Question 9

Question

A projectile is launched from ground level at an angle of 30◦above the horizon-

tal. The projectile lands 4.0 seconds later 120 meters downrange. Determine

the initial speed of the projectile.

Solution

Let’s denote the initial speed of the projectile as v, the horizontal component

of velocity as vx, and the vertical component of velocity as vy. We can use the

kinematic equations to solve for the initial speed.

Step 1: Identify the known values and convert the angle to radians. Given:



Angle of launch: θ= 30◦



Time of flight: t= 4.0 s



Range: x= 120 m



Acceleration due to gravity: g= 9.8 m/s2

Convert the angle to radians:

θrad =θ×π

180= 30◦×π

180=π

6radians

Step 2: Calculate the initial velocity components. The horizontal compo-

nent of velocity remains constant (vx=vcos θ), and the vertical component

of velocity will change due to gravity (vy=vsin θ−gt). The range can be

expressed as x=vxt.

Step 3: Find the horizontal and vertical component velocities. From the

range equation, we have:

vx=x

t=120 m

4.0 s = 30 m/s

Now, substitute vxinto the vertical component equation:

vy=vsin θ−gt

vsin θ=vy+gt

v=vy+gt

sin θ

Substitute the known values:

v=0 + (9.8 m/s2×4.0 s)

sin π

6

v=39.2 m/s

0.5

v= 78.4 m/s

Therefore, the initial speed of the projectile is 78.4 m/s .

Question 10

Question

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

with an initial speed of 30 m/s. What are the horizontal and vertical components

of the projectile’s velocity 2 s after it was launched?

Solution

Let’s first find the horizontal and vertical components of the projectile’s velocity

at 2 s after it was launched.

Step 1: Find the horizontal component of velocity (vx): The horizontal

component of the projectile’s velocity remains constant (no horizontal accel-

eration). Given that the initial velocity (v0) makes an angle of 60◦with the

horizontal, the horizontal component of velocity is given by:

vx=v0·cos(60◦) = 30 m/s ·cos(60◦) = 15 m/s

Step 2: Find the vertical component of velocity (vy): The vertical compo-

nent of the projectile’s velocity changes due to the acceleration due to gravity.

The vertical component of velocity at any time tafter launch is given by:

vy=v0·sin(60◦)−g·t

Where: - v0= 30 m/s is the initial speed, - g=−9.81 m/s2is the acceleration

due to gravity, and - t= 2 s is the time elapsed. Substitute the values to find

vy:

vy= 30 m/s ·sin(60◦)−9.81 m/s2·2 s = 25.98 m/s −19.62 m/s = 6.36 m/s

Therefore, 2 s after the projectile was launched, its horizontal component of

velocity is 15 m/s and the vertical component of velocity is 6.36 m/s.

Question 11

Question

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

the horizontal. Determine the baseball’s maximum height above the ground.

Assume air resistance is negligible and g= 9.81 m/s2.

Solution

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

The initial velocity (v0) of the baseball can be resolved into horizontal (v0x)

and vertical (v0y) components using trigonometric functions. Since the launch

angle is 30◦above the horizontal, we have:

v0x=v0cos(30◦) and v0y=v0sin(30◦)

Step 2: Determine the time it takes for the baseball to reach its maximum

height. At the maximum height, the vertical component of the velocity becomes

zero. We can use the kinematic equation v=v0+at for the vertical motion,

where v0is the initial vertical velocity, ais the acceleration due to gravity, and

tis the time taken. Since the final vertical velocity is 0 m/s at the maximum

height, we have:

0 = v0y−gt

Solving for t, we get:

t=v0y

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

mum height (H) reached by the baseball can be determined using the kinematic

equation:

H=v0yt−1

2gt2

Substitute the values of v0yand tinto the equation to find the maximum height.

H= (v0sin(30◦)) v0sin(30◦)

g−1

2gv0sin(30◦)

g2

H= (1

2v2

0sin2(30◦)) −1

0sin2(30◦)

g

H=1

2v2

0sin2(30◦)1−1

g

Simplify the expression to find the maximum height above the ground.

Question 12

Question

A projectile is launched with an initial speed of 20 m/s at an angle of 30 degrees

above the horizontal. Calculate the maximum height above the starting point

that the projectile reaches. Assume air resistance is negligible and take the

acceleration due to gravity as 9.81 m/s2.

Solution

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

v0= 20 m/s and θ= 30◦be the initial speed and angle of launch, respectively.

The horizontal component of velocity is given by: v0x=v0cos(θ), and the

vertical component of velocity is given by: v0y=v0sin(θ). Substitute the given

values to find v0xand v0y:v0x= 20 cos(30◦)≈17.32 m/s, v0y= 20 sin(30◦) =

10 m/s.

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

The time to reach maximum height is given by t=v0y

g. Substitute v0yand

g= 9.81 m/s2to find t:t=10 m/s

9.81 m/s2≈1.02 s.

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

imum height reached is given by h=v0yt−1

2gt2. Substitute v0y,t, and gto

find h:h= 10 m/s ×1.02 s −1

2×9.81 m/s2×(1.02 s)2≈5.1 m.

Therefore, the maximum height above the starting point that the projectile

reaches is approximately 5.1 meters.

Question 13

Question

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

above the horizontal. Determine the following:

1. The maximum height reached by the projectile (in meters).

2. The total time of flight (in seconds).

Assume the acceleration due to gravity is −9.81 m/s2.

Solution

Given: Initial velocity, vi= 30 m/s Launch angle, θ= 45◦Acceleration due to

gravity, g=−9.81 m/s2

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

nents. The initial horizontal velocity is given by vix =vicos θand the initial

vertical velocity is given by viy =visin θ.

vix = 30 cos 45◦

= 30 ×1

√2

= 15√2 m/s

viy = 30 sin 45◦

= 30 ×1

√2

= 15√2 m/s

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

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

use the kinematic equation vf=vi+at to find the time t.

0 = 15√2−9.81t

t=15√2

9.81

Question 14

Question

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

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

during its flight. Assume the acceleration due to gravity is 9.8 m/s2.

Solution

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

The initial speed of the projectile is 30 m/s and it is launched at an angle of 60

degrees above the horizontal. Therefore, the initial vertical component of the

velocity is given by:

vi,y =visin θ= 30 m/s ×sin(60◦)

vi,y = 30 m/s ×√3

2= 15√3 m/s

Step 2: Use kinematic equations to find the time taken to reach maximum

height. The time taken to reach maximum height can be found using the ver-

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

velocity is zero.

vf,y =vi,y +at

0 = 15√3−9.8t

t=15√3

9.8

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

imum height can be determined using the vertical displacement equation.

ymax =vi,yt−1

2gt2

ymax = 15√3×15√3

9.8−1

2×9.8 15√3

9.8!2

ymax =225 ×3

9.8−225 ×3

2×9.8

ymax =675

9.8−337.5

9.8

ymax ≈68.88 m

Therefore, the maximum height reached by the projectile is approximately

68.88 meters.

Question 15

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 it reaches and the total time

it takes for the projectile to hit the ground.

Solution

Step 1: Resolve the initial velocity into its x and y components. The initial

velocity v0can be resolved into its x-component v0xand y-component v0yusing

the following equations:

v0x=v0cos(θ)

v0y=v0sin(θ)

Substitute v0= 20 m/s and θ= 30◦:

v0x= 20 cos(30◦) = 20 ×√3

2= 10√3 m/s

v0y= 20 sin(30◦) = 20 ×1

2= 10 m/s

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

tile reaches the maximum height when its vertical velocity component becomes

zero. The time taken to reach the maximum height can be determined using

the following kinematic equation:

vyf =v0y−gt

Where vyf = 0 (at maximum height) and g= 9.81 m/s2. Substitute the known

values into the equation:

0 = 10 −9.81t

t=10

9.81 ≈1.02 s

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

be calculated using the following kinematic equation:

h=v0yt−1

2gt2

Substitute v0y= 10 m/s, t= 1.02 s, and g= 9.81 m/s2:

h= 10 ×1.02 −1

2×9.81 ×(1.02)2

h≈5.1−5.1≈0 m

Step 4: Calculate the total time of flight. The total time of flight is twice

the time taken to reach the maximum height, since the flight is symmetrical.

Therefore, the total time Tis:

T= 2t= 2 ×1.02

T≈2.04 s

Therefore, the maximum height reached by the projectile is 0 m and the

total time it takes to hit the ground is 2.04 s.

Question 16

Question

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

of 25 m/s. Calculate the total time the baseball is in the air and the maximum

height it reaches. Take g= 9.81 m/s2for the acceleration due to gravity.

Solution

Step 1: Resolve initial velocity into horizontal and vertical components: Let

v0= 25 m/s be the initial velocity and θ= 35◦be the launch angle. The initial

velocity components are:

v0x=v0cos θ

v0y=v0sin θ

Step 2: Calculate time to reach maximum height: The time to reach maxi-

mum height can be found by using the vertical component of the initial velocity

and acceleration due to gravity:

v0y=vy−gt

At maximum height, vy= 0:

0 = v0y−gt

t=v0y

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

baseball can be calculated using the equation for the vertical displacement at

maximum height:

y=v2

0y−2g∆y

At maximum height, vy= 0:

0 = v2

0y−2g∆y

∆y=v2

Step 4: Calculate total time in the air: The total time in the air is twice the

time to reach maximum height:

Total time = 2t

Substitute the given values to find the answers. - v0y= 25 ·sin 35◦-t=

25·sin 35◦

9.81 - ∆y=(25·sin 35◦)2

2·9.81 - Total time = 2 ·25·sin 35◦

9.81

Question 17

Question

A baseball is hit at an angle of 40◦above the horizontal with an initial speed of

25 m/s. Calculate the maximum height the baseball reaches during its flight.

Solution

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

components.

vix=vi·cos(θ)

viy=vi·sin(θ)

Step 2: Substitute the given values into the equations.

vix= 25 m/s ·cos(40◦)≈19.12 m/s

viy= 25 m/s ·sin(40◦)≈16.06 m/s

Step 3: Calculate the time taken for the baseball to reach its maximum

height using the vertical component of the initial velocity.

vf=viy+agravity ·t

Since the baseball reaches its maximum height, the final velocity is 0 m/s. So,

0 m/s = 16.06 m/s −9.81 m/s2·t

Step 4: Solve for the time t.

t=16.06 m/s

9.81 m/s2≈1.64 s

Step 5: Calculate the maximum height by using the equation for vertical

displacement when the final velocity is zero.

ymax =viy·t−1

2agravity ·t2

ymax = 16.06 m/s ·1.64 s −1

2·9.81 m/s2·(1.64 s)2

ymax ≈13.2 m

Therefore, the maximum height the baseball reaches during its flight is ap-

proximately 13.2 meters.

Question 18

Question

A particle moves in the xy plane with a position vector r = 3t2ˆ

i+(5−2t)ˆ

j, where

r is in meters and tis in seconds. Find the particle’s velocity and acceleration

as functions of t.

Solution

Step 1: To find the particle’s velocity, we differentiate the position vector r with

respect to time t.

Velocity v =dr

dt(3t2ˆ

i+ (5 −2t)ˆ

= 6tˆ

i−2ˆ

Step 2: To find the particle’s acceleration, we differentiate the velocity vector

v with respect to time t.

Acceleration a =dv

dt(6tˆ

i−2ˆ

= 6ˆ

Therefore, the particle’s velocity is v = 6tˆ

i−2ˆ

jm/s and the acceleration is

a = 6ˆ

im/s2as functions of time t.

Question 19

Question

A particle moves in the x-yplane with an acceleration given by a = (3t)ˆ

i−(4)ˆ

where ˆ

iand ˆ

jare unit vectors in the xand ydirections, respectively. If at t= 0

the particle is at the origin with a velocity of v0= (5 m/s)ˆ

i+ (2 m/s)ˆ

j, find the

velocity vector as a function of time.

Solution

Step 1: Integrate the acceleration vector a to get the velocity vector v.

Given: a = (3t)ˆ

i−(4)ˆ

Integrating each component with respect to time: Raxdt =R3t dt and

Raydt =R−4dt

We get: vx=R3t dt =1

2(3t2) = 3

2t2+Cx

vy=R−4dt =−4t+Cy

Step 2: Apply the initial conditions to find the constants Cxand Cy.

Given initial velocity v0= (5 m/s)ˆ

i+(2 m/s)ˆ

j, which means at t= 0: vx(0) =

5 and vy(0) = 2

Using these initial conditions: vx(0) = 3

2(0)2+Cx=Cx= 5 vy(0) =

−4(0) + Cy=Cy= 2

So, vx=3

2t2+ 5 and vy=−4t+ 2

Therefore, the velocity vector as a function of time is: v =3

2t2+ 5ˆ

(−4t+ 2)ˆ

Question 20

Question

A particle moves along a curved path defined by the parametric equations x(t) =

3t2and y(t)=4t3. Find the magnitude of the acceleration of the particle at

t= 2 seconds.

Solution

Step 1: Find the velocity vector v of the particle at t= 2 seconds by taking the

derivative of the position vector r(t) = x(t)ˆ

i+y(t)ˆ

jwith respect to time:

v =dr

dt =d

dt(3t2ˆ

i+ 4t3ˆ

j)=6tˆ

i+ 12t2ˆ

Now, evaluate v at t= 2 seconds:

v(2) = 6(2)ˆ

i+ 12(2)2ˆ

j= 12ˆ

i+ 48ˆ

Step 2: Find the acceleration vector a of the particle at t= 2 seconds by

taking the derivative of the velocity vector v with respect to time:

a =dv

dt =d

dt(6tˆ

i+ 12t2ˆ

j) = 6ˆ

i+ 24tˆ

Now, evaluate a at t= 2 seconds:

a(2) = 6ˆ

i+ 24(2)ˆ

j= 6ˆ

i+ 48ˆ

Step 3: Calculate the magnitude of the acceleration vector a at t= 2 seconds:

|a(2)|=p(6)2+ (48)2=√36 + 2304 = √2340 ≈48.38 m/s2

Therefore, the magnitude of the acceleration of the particle at t= 2 seconds

is approximately 48.38 m/s2.

Question 21

Question

A particle moves in the xy plane with an acceleration given by a = (4t2ˆ

i−

6tˆ

j) m/s2. If the particle starts from rest at the origin, find its velocity vector

as a function of time.

Solution

Step 1: To find the velocity vector as a function of time, we need to integrate the

acceleration function with respect to time. Step 2: Integrating the xcomponent

of acceleration, ax= 4t2m/s2, with respect to time, we get the xcomponent of

velocity:

Zaxdt =Z4t2dt =4

3t3+C1

where C1is the constant of integration. The xcomponent of velocity is vx=

3t3+C1. Step 3: Integrating the ycomponent of acceleration, ay=−6tm/s2,

with respect to time, we get the ycomponent of velocity:

Zaydt =Z−6t dt =−3t2+C2

where C2is the constant of integration. The ycomponent of velocity is vy=

−3t2+C2. Step 4: The velocity vector is given by v =vxˆ

i+vyˆ

j. Substituting

in the expressions for vxand vy, we have the velocity vector as a function of

time:

v =4

3t3+C1ˆ

i+−3t2+C2ˆ

where C1and C2are constants of integration.

Question 22

Question

A soccer player kicks the ball from the ground at an angle of 40◦above the

horizontal with an initial speed of 20 m/s. Find the maximum height the ball

reaches and the total time it is in the air before hitting the ground. Neglect air

resistance.

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 component

v0xand vertical component v0y:

v0x=v0cos(40◦)

v0y=v0sin(40◦)

Step 2: Find the time the ball takes to reach the maximum height. At

the maximum height, the vertical component of the velocity is 0 m/s. Using

the equation of motion vy=v0y−gt, where v0yis the initial vertical velocity,

g= 9.8 m/s2is the acceleration due to gravity, and tis the time taken to reach

the maximum height:

0 = v0y−gt

t=v0y

Step 3: Calculate the maximum height reached by the ball. Using the equa-

tion of motion y=y0+v0yt−1

2gt2, where y0= 0 m is the initial vertical

position, v0yis the initial vertical velocity, g= 9.8 m/s2is the acceleration due

to gravity, and tis the time taken to reach the maximum height:

ymax =v0y×v0y

g−1

2gv0y

g2

Step 4: Find the total time the ball is in the air. The total time the ball

is in the air is twice the time it takes to reach the maximum height since the

vertical motion is symmetrical.

Total time = 2 ×v0y

Question 23

Question

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

horizontal. Determine the maximum height the ball reaches and the total time

of flight.

Solution

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

Let the initial velocity vi= 20 m/s, the angle above the horizontal θ= 30◦. The

horizontal component vix=vicos θand the vertical component viy=visin θ.

vix= 20 m/s cos 30◦= 20 m/s ×√3

2= 10√3 m/s

viy= 20 m/s sin 30◦= 20 m/s ×1

2= 10 m/s

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

The time tit takes for the ball to reach the maximum height can be found using

the vertical motion equation y=viyt+1

2ayt2. Since the ball reaches maximum

height when vy= 0, we have 0 = viy−gtmax →tmax =viy

Substitute viy= 10 m/s and g= 9.8 m/s2:

tmax =10

9.8≈1.02 s

Step 3: Calculate the maximum height the ball reaches.

The maximum height hmax can be calculated using the vertical motion equation

y=v2

iy−2ayy. Since the final vertical velocity vy= 0 at the maximum height,

we have 0 = v2

iy−2ghmax.

Solve for hmax:

hmax =v2

2g=(10 m/s)2

2×9.8 m/s2≈5.10 m

Step 4: Determine the total time of flight.

Since the ball is thrown symmetrically, the total time of flight Tis twice the

time to reach maximum height, T= 2 ×tmax.

Substitute tmax = 1.02 s into the equation:

T= 2 ×1.02 s = 2.04 s

Therefore, the maximum height the ball reaches is approximately 5.10 m and

the total time of flight is 2.04 s.

Question 24

Question

A particle moves along the following path in the xy plane: it starts at the origin,

moves to the point (3 m,4 m), then to the point (−2 m,1 m), and finally ends

at the point (−1 m,−5 m). Calculate the total distance traveled by the particle.

Solution

Step 1: Calculate the distance between the initial and final points.

The distance between two points (x1, y1) and (x2, y2) in the xy plane is given

by the formula:

D=p(x2−x1)2+ (y2−y1)2

Using this formula, we can calculate the distances between the sequential points:

- From (0,0) to (3,4):

D1=p(3 −0)2+ (4 −0)2=p32+ 42= 5 m

- From (3,4) to (−2,1):

D2=p(−2−3)2+ (1 −4)2=p(−5)2+ (−3)2=√25 + 9 = √34 m

- From (−2,1) to (−1,−5):

D3=p(−1 + 2)2+ (−5−1)2=√1 + 36 = √37 m

Step 2: Calculate the total distance traveled.

The total distance is the sum of the individual distances:

Total distance = D1+D2+D3= 5 m+√34 m+√37 m ≈5 m+5.83 m+6.08 m ≈16.91 m

Therefore, the total distance traveled by the particle is approximately 16.91

meters.

Question 25

Question

A baseball player hits a ball at an angle of 35◦above the horizontal. The ball

leaves the bat with a speed of 30 m/s and an initial height of 1.5 m above the

ground. How far does the ball travel horizontally before hitting the ground?

Solution

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

Step 2: Find the time of flight of the ball. Step 3: Calculate the horizontal

distance the ball travels before hitting the ground.

Step 1: Resolve the initial velocity: The initial velocity of the ball can be

resolved into horizontal (V0x) and vertical (V0y) components using trigonometry:

V0x=V0cos θand V0y=V0sin θ

Substitute V0= 30 m/s and θ= 35◦into the above equations:

V0x= 30 cos 35◦≈24.65 m/s

V0y= 30 sin 35◦≈17.12 m/s

Step 2: Find the time of flight: The time of flight (t) can be found using

the equation for vertical motion:

y=V0yt+1

2gt2

where yis the initial height (1.5 m) and gis the acceleration due to gravity

(-9.81 m/s2). Since the ball lands back on the ground, the final vertical position

is 0. Solving for t:

0 = 17.12t+1

2(−9.81)t2

−4.905t2+ 17.12t= 0

t(17.12 −4.905t)=0

t= 0 or t=17.12

4.905 ≈3.49 s

Step 3: Calculate the horizontal distance: The horizontal distance the ball

travels can be calculated using the equation for horizontal motion:

x=V0xt

Substitute V0x= 24.65 m/s and t= 3.49 s into the above equation:

x= 24.65 ×3.49 ≈86.00 m

Therefore, the ball travels approximately 86.00 meters horizontally before hit-

ting the ground.

Question 26

Question

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

horizontal. Calculate the maximum height reached by the ball. Assume the

acceleration due to gravity is 9.8 m/s2.

Solution

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

The initial vertical component of velocity is: viy=visin θ= 20 m/s ×sin(30◦)

Step 2: Calculate the initial vertical component of velocity. viy= 20 m/s ×

2= 10 m/s

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

height. The equation for vertical motion is: y=yi+viyt−1

2gt2

At the maximum height, the final vertical velocity is zero. Therefore, vfy =

Step 4: Find the time taken to reach the maximum height. 0 = viy −gt

t=viy

Step 5: Substitute the values to find the maximum height. y= 0+(10 m/s)×

10 m/s

9.8 m/s2−1

2×9.8 m/s2×10 m/s

9.8 m/s22

Step 6: Calculate the maximum height. y= 0 + 10 m −1

2×9.8 m/s2×10

9.8

y= 10 m −5 m = 5 m

Therefore, the maximum height reached by the ball is 5 m.

Question 27

Question

A baseball player hits a ball with an initial velocity of 40 m/s at an angle of 30

degrees above the horizontal. Determine the maximum height the ball reaches

during its flight.

Solution

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

The horizontal component of the velocity is given by v0x=v0·cos(θ) and the

vertical component is given by v0y=v0·sin(θ). Given that v0= 40 m/s and

θ= 30◦, we have: v0x= 40·cos(30◦) = 40·√3

2= 20√3 m/s v0y= 40 ·sin(30◦) =

40 ·1

2= 20 m/s

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

The time it takes to reach the maximum height can be found using the vertical

component of the velocity and the acceleration due to gravity. The formula is

vf=vi+at, where vf= 0 at the maximum height. We have: 0 = v0y−gt

Solving for t, we get: t=v0y

g=20

9.8≈2.04 s

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

using the vertical motion equation y=v0yt−1

2gt2. Substitute the values we

obtained: y= 20 ·2.04 −1

2·9.8·(2.04)2Final calculation gives: y≈20.4 m

Therefore, the maximum height the ball reaches is approximately 20.4 me-

ters.

Question 28

Question

A hockey puck is initially at rest on the ice. It is then struck by a hockey

stick, giving it an initial velocity of 10 m/s at an angle of 30 degrees above the

horizontal. The coefficient of kinetic friction between the puck and the ice is

0.1. How far does the puck travel before coming to rest?

Solution

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

The initial velocity of the puck is 10 m/s at an angle of 30 degrees above

the horizontal. Let vixrepresent the initial velocity in the x-direction and viy

represent the initial velocity in the y-direction.

vix= 10 cos(30◦) = 8.66 m/s

viy= 10 sin(30◦) = 5 m/s

Step 2: Determine the acceleration of the puck in the x-direction. Since the

puck is on ice, the frictional force fk=µk·Nwill act against the motion of

the puck in the x-direction. The normal force Ncan be calculated using the

gravitational force.

N=mg

N= 0.1·9.8 m/s2= 0.98 N

fk= 0.1·0.98 = 0.098 N

ax=−fk

m=−0.098

1=−0.098 m/s2

Step 3: Determine the time for the puck to stop in the x-direction. The final

velocity in the x-direction is 0 m/s since the puck eventually comes to a stop.

vfx=vix+axt

0=8.66 −0.098t

t=8.66

0.098 = 88.3 s

Step 4: Calculate the distance traveled by the puck before coming to rest.

The distance traveled in the x-direction can be calculated using the equation:

dx=vixt+1

2axt2

dx= 8.66(88.3) + 0.5(−0.098)(88.3)2

dx= 763.7 m

Therefore, the puck travels 763.7 meters before coming to a stop.

Question 29

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 reached by the projectile.

Solution

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

Given: Initial speed, v0= 30 m/s Launch angle, θ= 60◦

The horizontal component of the initial velocity is:

v0x=v0cos θ

v0x= 30 cos 60◦

v0x= 30 ×1

v0x= 15 m/s

The vertical component of the initial velocity is:

v0y=v0sin θ

v0y= 30 sin 60◦

v0y= 30 ×√3

v0y= 15√3 m/s

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

to reach maximum height. The equation for vertical motion is:

y=v0yt−1

2gt2

At maximum height, the vertical velocity is zero. So, vy= 0 Hence, we can use

the equation:

0 = v0y−gt

Solving for t:

15√3=9.8t

t≈15√3

9.8

t≈2.73 sec

Step 3: Find the height attained by the projectile. Using the displacement

formula in the vertical direction:

y=v0yt−1

2gt2

Substitute the known values:

y= 15√3×2.73 −1

2×9.8×(2.73)2

y≈30.88 m

Therefore, the maximum height reached by the projectile is approximately

30.88 meters.

Question 30

Question

A cannonball is fired with an initial speed of 100 m/s at an angle of 30◦above

the horizontal. Determine the maximum height the cannonball reaches and the

total time it is in the air.

Solution

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

Given that the initial speed is 100 m/s and the angle of projection is 30◦, the

horizontal component of velocity (v0x) and the vertical component of velocity

(v0y) can be calculated as follows:

v0x=v0cos(30◦)

v0y=v0sin(30◦)

Step 2: Calculate the time of flight. Using the kinematic equation y=

v0yt+1

2at2, we can find the time it takes for the cannonball to reach its max-

imum height. At the maximum height, the vertical component of velocity is 0.

Therefore:

0 = v0y−gt

t=v0y

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

using the kinematic equation y=v0yt−1

2gt2. Substituting the time of flight

calculated in Step 2:

y=v0y·v0y

g−1

2g·v0y

g2

Step 4: Calculate the total time in the air. As the total time in the air is

twice the time of flight, the total time in the air is:

Total time = 2t

Question 31

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 reached by the projectile.

Solution

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

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

components as follows: The horizontal component: v0x=v0cos(θ) Given that

v0= 30 m/s and θ= 60◦, we have: v0x= 30 cos(60◦) = 30 ×1

2= 15 m/s

The vertical component: v0y=v0sin(θ)v0y= 30 sin(60◦) = 30×√3

2= 15√3

m/s

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

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

initial velocity and acceleration due to gravity. The equation relating initial

velocity, acceleration, time, and displacement in the vertical direction is: vy=

v0y−gt, where vy= 0 at maximum height. Solving for tgives: t=v0y

15√3

9.81 ≈1.54 s

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

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

component of the initial velocity, time to reach maximum height, and accelera-

tion due to gravity. The equation for vertical displacement is: h=v0yt−1

2gt2

Substitute v0y= 15√3 m/s, t≈1.54 s, and g= 9.81 m/s2into the above equa-

tion to find the maximum height: h= 15√3×1.54 −1

2×9.81 ×(1.54)2≈23.83

Therefore, the maximum height reached by the projectile is approximately

23.83 meters.

Question 32

Question

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

an angle of 60 degrees above the horizontal. Find the projectile’s total time of

flight and maximum height reached. Take the acceleration due to gravity as

−9.81 m/s2.

Solution

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



Let the initial velocity v0= 30 m/s and the angle of elevation θ= 60◦.



The horizontal component of the initial velocity v0x=v0cos θ.



The vertical component of the initial velocity v0y=v0sin θ.



Substituting v0= 30 m/s and θ= 60◦into the above formulas: v0x=

30 cos 60◦and v0y= 30 sin 60◦.

Step 2: Calculate the time of flight using the vertical component of motion.



The time of flight in projectile motion can be determined by considering

the vertical component.



The formula vf=v0+at can be used to find the time taken to reach

maximum height.



At the peak of the projectile’s trajectory, the vertical component of the

final velocity vfy = 0.



Substituting vfy = 0, v0y= 30 sin 60◦, and a=−9.81 m/s2into the

formula: 0 = 30 sin 60◦−9.81t. Solve for t.

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



The maximum height can be calculated using the vertical motion equation

y=v0yt+1

2at2.



The time tfound in Step 2 can be used to calculate the maximum height

ymax.



Substituting v0y= 30 sin 60◦,tfrom Step 2, and a=−9.81 m/s2into the

above equation.

Question 33

Question

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

of 20 m/s from the edge of a cliff that is 50 m high. How far from the base of

the cliff does the projectile strike the ground?

Solution

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

The initial velocity (v0) of the projectile can be resolved into its horizontal

component (v0x) and vertical component (v0y) as follows:

v0x=v0cos θ

v0y=v0sin θ

where v0= 20 m/s and θ= 30◦.

Step 2: Calculate the time taken for the projectile to hit the ground. The

time taken for the projectile to hit the ground can be found using the vertical

motion equation:

y=v0yt+1

2at2

where y= -50 m (height of the cliff), v0y=v0sin θ,a= -9.8 m/s2(acceleration

due to gravity), and tis the time of flight. Solving for tgives:

−50 = (20 sin 30◦)t−1

2(9.8)t2

Step 3: Find the horizontal distance traveled by the projectile. The horizon-

tal distance traveled by the projectile can be determined using the horizontal

motion equation:

x=v0xt

where v0x=v0cos θ. Substituting the known values into the equation gives:

x= (20 cos 30◦)t

Step 4: Calculate the horizontal distance from the base of the cliff. Finally,

substitute the value of tfrom Step 2 into the equation from Step 3 to find the

horizontal distance:

x= (20 cos 30◦)t

Question 34

Question

A projectile is launched 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

projectile.

Solution

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

The initial velocity can be resolved into two components: - The horizontal

component vix=vicos(θ), - The vertical component viy=visin(θ),

where viis the initial velocity (20 m/s) and θis the launch angle (30◦).

Calculating the components: vix= 20 cos(30◦) = 20 ·√3

2= 10√3 m/s, viy=

20 sin(30◦) = 20 ·1

2= 10 m/s.

Step 2: Determine the time it takes for the projectile to reach the maximum

height. Since the projectile reaches maximum height when the vertical compo-

nent of velocity is zero, we can use the equation vf=vi−gt, where vfis the

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

Substitute into the equation: 0 = 10 −9.8t, Solving for tgives: t=10

9.8= 1.02

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

found using the equation: h=viyt−1

2gt2. Substitute the known values: h=

10 ·1.02 −1

2·9.8·(1.02)2= 10.2−5·1.02 = 10.2−5.1 = 5.1 m.

Therefore, the maximum height reached by the projectile is 5.1 meters.

Question 35

Question

A stone is thrown at an angle of 30◦above the horizontal with an initial speed

of 20 m/s from the edge of a cliff 80 m high. Find the total time the stone is in

the air.

Solution

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

components. The horizontal component of the initial velocity V0xis given by:

V0x=V0cos(θ)

where V0is the initial speed and θis the angle of projection. Substitute V0=

20 m/s and θ= 30◦into the equation:

V0x= 20 cos(30◦)

V0x= 20 ·√3

V0x= 10√3 m/s

The vertical component of the initial velocity V0yis given by:

V0y=V0sin(θ)

Substitute V0= 20 m/s and θ= 30◦into the equation:

V0y= 20 sin(30◦)

V0y= 20 ·1

V0y= 10 m/s

Step 2: Determine the time taken for the stone to reach the peak of its

trajectory. Use the vertical motion equation: y=V0yt−1

2gt2. At the peak

of the trajectory, the stone reaches a height of 80 m, so y= 80 m. Substitute

V0y= 10 m/s and g= 9.81 m/s2into the equation:

80 = 10t−1

2·9.81 ·t2

Step 3: Solve for the time taken to reach the peak of the trajectory. Rear-

range the equation to get a quadratic equation in the form at2+bt +c= 0:

−4.905t2+ 10t−80 = 0

Question 2

Question

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

speed of 20 m/s. Determine the projectile’s maximum height above its launch

point.

Solution

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

The initial velocity v0can be resolved into its horizontal and vertical components

as:

v0x=v0cos(θ) = 20 m/s ·cos(30◦) = 17.32 m/s

v0y=v0sin(θ) = 20 m/s ·sin(30◦) = 10 m/s

Step 2: Use the vertical component of velocity to find the time to reach

maximum height. The time to reach maximum height can be found using the

vertical component of velocity:

vy=v0y−gt

where g= 9.8 m/s2is the acceleration due to gravity. At maximum height, the

vertical component of velocity is zero, so:

0 = 10 m/s −9.8 m/s2·t

Solving for t, we get:

t=10 m/s

9.8 m/s2= 1.02 s

Step 3: Calculate the maximum height. The maximum height (ymax) above

the launch point can be found using the kinematic equation:

ymax =v0yt−1

2gt2

Substitute the known values:

ymax = 10 m/s ·1.02 s −1

2·9.8 m/s2·(1.02 s)2

ymax = 10.20 m −5.10 m = 5.10 m

Therefore, the projectile’s maximum height above its launch point is 5.10

meters.

Question 3

Question

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

and lands 4 seconds later. If air resistance is neglected, determine the initial

speed of the projectile.

Solution

Step 1: Identify the known values and equations related to projectile motion.

In projectile motion, we can break the initial velocity into its horizontal (v0x)

and vertical (v0y) components. The key equations for projectile motion in two

dimensions are: 1. vx=v0x2. vy=v0y−gt 3. x=v0xt4. y=v0yt−1

2gt25.

y= (v0y−gt)2=v2

0y−2gy

Step 2: Determine the initial velocity components using trigonometric rela-

tionships.

Given that the projectile is launched at an angle of 30◦above the horizontal,

we can express the initial velocity as: v0x=v0cos(30◦)v0y=v0sin(30◦)

Step 3: Use the equation for time of flight to find the initial velocity.

The time of flight of the projectile is given as 4 seconds. This means the

projectile spends 4 seconds in the air, so t= 4s. Using the equation for

time: y=v0yt−1

2gt20 = v0yt−1

2gt2v0y=1

2gt Substitute v0y=1

2gt into

v0x=v0cos(30◦) and v0y=v0sin(30◦). vx=v0cos(30◦)vy=1

2g(4s)

Step 4: Calculate the initial speed of the projectile.

To find the initial speed, we need to find v0by summing the squares of the veloci-

ties in the x and y components: v0=qv2

x+v2

yv0=q(v0cos(30◦))2+1

2g(4s)2

Solve for v0.

Question 4

Question

A particle moves in the xy plane. Its position vector is given by r = (3t2−

2t)ˆ

i+ (4t+ 1)ˆ

j, where tis in seconds, xand yare in meters. Find the velocity

and acceleration vectors of the particle.

Solution

Step 1: To find the velocity vector v, we differentiate the position vector r with

respect to time t.

v =dr

Step 2: Differentiating each component of r with respect to t:

dt =d(3t2−2t)

dt = 6t−2

dt =d(4t+ 1)

dt = 4

Step 3: Therefore, the velocity vector v is:

v = (6t−2)ˆ

i+ 4ˆ

Step 4: To find the acceleration vector a, we differentiate the velocity vector

v with respect to time t.

a =dv

Step 5: Differentiating each component of v with respect to t:

d2x

dt2=d(6t−2)

dt = 6

d2y

dt2=d(4)

dt = 0

Step 6: Therefore, the acceleration vector a is:

a = 6ˆ

Question 5

Question

A particle moves in the xy plane with an acceleration given by a= (4t2ˆ

i−

6ˆ

j) m/s2, where tis in seconds. The particle’s velocity at t= 2.0 s is v=

(5.0ˆ

i−5.0ˆ

j) m/s. Find the particle’s position vector at t= 2.0 s if the particle

is at the origin when t= 0 s.

Solution

Step 1: To find the particle’s position vector at t= 2.0 s, we first need to find

the particle’s velocity and position functions in terms of time.

Given the acceleration a= (4t2ˆ

i−6ˆ

j) m/s2, we can find the velocity function

by integrating the acceleration function with respect to time:

v=Zadt =Z4t2dtˆ

i−Z6dtˆ

v=4

3t3+C1ˆ

i−6t+C2ˆ

Given that the velocity at t= 2.0 s is v= (5.0ˆ

i−5.0ˆ

j) m/s, plug in t= 2.0 s

to find the constants C1and C2:

(5.0ˆ

i−5.0ˆ

j) = 4

3(2.0)3+C1ˆ

i−6(2.0) + C2ˆ

(5.0ˆ

i−5.0ˆ

j) = (10.67 + C1)ˆ

i−12 + C2ˆ

Therefore, C1=−0.67 m/s in the ˆ

idirection and C2= 7.0 m in the ˆ

jdirection.

The velocity function is then:

v=4

3t3−0.67ˆ

i−6t+ 7ˆ

Step 2: To find the position function, integrate the velocity function with

respect to time:

r=Zvdt =Z4

3t3−0.67dtˆ

i−Z6t+ 7 dtˆ

r=1

3t4−0.67t+C3ˆ

i−3t2+ 7t+C4ˆ

Since the particle is at the origin when t= 0 s, we know that C3= 0 and

C4= 0.

Thus, the position function is:

r=1

3t4−0.67tˆ

i−3t2+ 7tˆ

Finally, plug in t= 2.0 s to find the position vector at t= 2.0 s:

r(2.0) = 1

3(2.0)4−0.67(2.0)

Question 6

Question

A cannonball is fired with an initial velocity of 200 m/s at an angle of 30◦above

the horizontal. Find the maximum height it reaches and the time taken to reach

that height.

Solution

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

The horizontal component of velocity, Vix, is given by: Vix=Vi·cos(θ) =

200 ·cos(30◦) = 173.21 m/s.

The vertical component of velocity, Viy, is given by: Viy=Vi·sin(θ) =

200 ·sin(30◦) = 100 m/s.

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

vertical motion equation vf=vi+at, where vf= 0 at the maximum height

and a=−9.81 m/s2: 0 = 100 −9.81t

t=100

9.81 = 10.20 s.

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

vertical motion equation h=vit+1

2at2:h= 100 ·10.20 + 1

2(−9.81)(10.20)2

h= 1020 −499.59 = 520.41 m.

Therefore, the maximum height reached by the cannonball is 520.41 m and

the time taken to reach that height is 10.20 s.

Question 7

Question

A baseball player hits a ball with an initial velocity of 30 m/s at an angle of 45◦

above the horizontal. The outfield wall is 100 m away and 5 m high. Will the

ball clear the wall? Assume there is no air resistance.

Solution

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

initial horizontal velocity v0xis given by:

v0x=v0cos θ= 30 m/s ×cos 45◦= 21.2 m/s

The initial vertical velocity v0yis given by:

v0y=v0sin θ= 30 m/s ×sin 45◦= 21.2 m/s

Step 2: Calculate the time of flight. Use the equation:

time = 2v0y

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

time = 2×21.2

9.81 = 4.33 s

Step 3: Calculate the maximum height reached by the ball. Use the equation:

hmax =v2

0y/2g

hmax = (21.2)2/(2 ×9.81) = 22.8 m

Step 4: Calculate the horizontal distance the ball travels. Using the equation:

d=v0x×time

d= 21.2×4.33 = 91.9 m

Step 5: Check if the ball clears the wall. Since the wall is 5 m high and the

ball reaches a maximum height of 22.8 m, the ball easily clears the 5 m wall.

Therefore, the ball will clear the 5 m high wall and reach a maximum height

of 22.8 m.

Question 8

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 what height above the point of projection

is the projectile 2 seconds after being launched?

Solution

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

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

components as follows: The initial horizontal component is vix=vicos(θ) where

viis the initial speed of the projectile and θis the launch angle. The initial

vertical component is viy=visin(θ).

Step 2: Determine the time taken for the projectile to reach a height of 2

m. The equation that relates the vertical displacement, initial vertical velocity,

time, and acceleration is:

y=viyt−1

2gt2

Since the projectile is at a height of 2 m, y= 2 m. Substitute viy= 30 sin(60◦) =

15√3 m/s and g= 9.8 m/s2into the equation.

2 = 15√3t−1

2·9.8·t2

0 = 15√3t−4.9t2

Solving for t, we get t= 0 (extraneous) or t≈0.349 s.

Step 3: Calculate the height above the point of projection after 2 seconds.

The equation for vertical position at time tcan be given by:

y=viyt−1

2gt2

At t= 2 s, substitute viy= 15√3 m/s, t= 2 s, and g= 9.8 m/s2into the

equation.

y= 15√3·2−1

2·9.8·22

y= 30√3−19.6

y≈37.38 m

Therefore, the projectile is approximately 37.38 m above the point of pro-

jection after 2 seconds.

Question 9

Question

A projectile is launched from ground level at an angle of 30◦above the horizon-

tal. The projectile lands 4.0 seconds later 120 meters downrange. Determine

the initial speed of the projectile.

Solution

Let’s denote the initial speed of the projectile as v, the horizontal component

of velocity as vx, and the vertical component of velocity as vy. We can use the

kinematic equations to solve for the initial speed.

Step 1: Identify the known values and convert the angle to radians. Given:



Angle of launch: θ= 30◦



Time of flight: t= 4.0 s



Range: x= 120 m



Acceleration due to gravity: g= 9.8 m/s2

Convert the angle to radians:

θrad =θ×π

180= 30◦×π

180=π

6radians

Step 2: Calculate the initial velocity components. The horizontal compo-

nent of velocity remains constant (vx=vcos θ), and the vertical component

of velocity will change due to gravity (vy=vsin θ−gt). The range can be

expressed as x=vxt.

Step 3: Find the horizontal and vertical component velocities. From the

range equation, we have:

vx=x

t=120 m

4.0 s = 30 m/s

Now, substitute vxinto the vertical component equation:

vy=vsin θ−gt

vsin θ=vy+gt

v=vy+gt

sin θ

Substitute the known values:

v=0 + (9.8 m/s2×4.0 s)

sin π

6

v=39.2 m/s

0.5

v= 78.4 m/s

Therefore, the initial speed of the projectile is 78.4 m/s .

Question 10

Question

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

with an initial speed of 30 m/s. What are the horizontal and vertical components

of the projectile’s velocity 2 s after it was launched?

Solution

Let’s first find the horizontal and vertical components of the projectile’s velocity

at 2 s after it was launched.

Step 1: Find the horizontal component of velocity (vx): The horizontal

component of the projectile’s velocity remains constant (no horizontal accel-

eration). Given that the initial velocity (v0) makes an angle of 60◦with the

horizontal, the horizontal component of velocity is given by:

vx=v0·cos(60◦) = 30 m/s ·cos(60◦) = 15 m/s

Step 2: Find the vertical component of velocity (vy): The vertical compo-

nent of the projectile’s velocity changes due to the acceleration due to gravity.

The vertical component of velocity at any time tafter launch is given by:

vy=v0·sin(60◦)−g·t

Where: - v0= 30 m/s is the initial speed, - g=−9.81 m/s2is the acceleration

due to gravity, and - t= 2 s is the time elapsed. Substitute the values to find

vy:

vy= 30 m/s ·sin(60◦)−9.81 m/s2·2 s = 25.98 m/s −19.62 m/s = 6.36 m/s

Therefore, 2 s after the projectile was launched, its horizontal component of

velocity is 15 m/s and the vertical component of velocity is 6.36 m/s.

Question 11

Question

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

the horizontal. Determine the baseball’s maximum height above the ground.

Assume air resistance is negligible and g= 9.81 m/s2.

Solution

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

The initial velocity (v0) of the baseball can be resolved into horizontal (v0x)

and vertical (v0y) components using trigonometric functions. Since the launch

angle is 30◦above the horizontal, we have:

v0x=v0cos(30◦) and v0y=v0sin(30◦)

Step 2: Determine the time it takes for the baseball to reach its maximum

height. At the maximum height, the vertical component of the velocity becomes

zero. We can use the kinematic equation v=v0+at for the vertical motion,

where v0is the initial vertical velocity, ais the acceleration due to gravity, and

tis the time taken. Since the final vertical velocity is 0 m/s at the maximum

height, we have:

0 = v0y−gt

Solving for t, we get:

t=v0y

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

mum height (H) reached by the baseball can be determined using the kinematic

equation:

H=v0yt−1

2gt2

Substitute the values of v0yand tinto the equation to find the maximum height.

H= (v0sin(30◦)) v0sin(30◦)

g−1

2gv0sin(30◦)

g2

H= (1

2v2

0sin2(30◦)) −1

0sin2(30◦)

g

H=1

2v2

0sin2(30◦)1−1

g

Simplify the expression to find the maximum height above the ground.

Question 12

Question

A projectile is launched with an initial speed of 20 m/s at an angle of 30 degrees

above the horizontal. Calculate the maximum height above the starting point

that the projectile reaches. Assume air resistance is negligible and take the

acceleration due to gravity as 9.81 m/s2.

Solution

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

v0= 20 m/s and θ= 30◦be the initial speed and angle of launch, respectively.

The horizontal component of velocity is given by: v0x=v0cos(θ), and the

vertical component of velocity is given by: v0y=v0sin(θ). Substitute the given

values to find v0xand v0y:v0x= 20 cos(30◦)≈17.32 m/s, v0y= 20 sin(30◦) =

10 m/s.

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

The time to reach maximum height is given by t=v0y

g. Substitute v0yand

g= 9.81 m/s2to find t:t=10 m/s

9.81 m/s2≈1.02 s.

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

imum height reached is given by h=v0yt−1

2gt2. Substitute v0y,t, and gto

find h:h= 10 m/s ×1.02 s −1

2×9.81 m/s2×(1.02 s)2≈5.1 m.

Therefore, the maximum height above the starting point that the projectile

reaches is approximately 5.1 meters.

Question 13

Question

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

above the horizontal. Determine the following:

1. The maximum height reached by the projectile (in meters).

2. The total time of flight (in seconds).

Assume the acceleration due to gravity is −9.81 m/s2.

Solution

Given: Initial velocity, vi= 30 m/s Launch angle, θ= 45◦Acceleration due to

gravity, g=−9.81 m/s2

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

nents. The initial horizontal velocity is given by vix =vicos θand the initial

vertical velocity is given by viy =visin θ.

vix = 30 cos 45◦

= 30 ×1

√2

= 15√2 m/s

viy = 30 sin 45◦

= 30 ×1

√2

= 15√2 m/s

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

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

use the kinematic equation vf=vi+at to find the time t.

0 = 15√2−9.81t

t=15√2

9.81

Question 14

Question

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

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

during its flight. Assume the acceleration due to gravity is 9.8 m/s2.

Solution

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

The initial speed of the projectile is 30 m/s and it is launched at an angle of 60

degrees above the horizontal. Therefore, the initial vertical component of the

velocity is given by:

vi,y =visin θ= 30 m/s ×sin(60◦)

vi,y = 30 m/s ×√3

2= 15√3 m/s

Step 2: Use kinematic equations to find the time taken to reach maximum

height. The time taken to reach maximum height can be found using the ver-

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

velocity is zero.

vf,y =vi,y +at

0 = 15√3−9.8t

t=15√3

9.8

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

imum height can be determined using the vertical displacement equation.

ymax =vi,yt−1

2gt2

ymax = 15√3×15√3

9.8−1

2×9.8 15√3

9.8!2

ymax =225 ×3

9.8−225 ×3

2×9.8

ymax =675

9.8−337.5

9.8

ymax ≈68.88 m

Therefore, the maximum height reached by the projectile is approximately

68.88 meters.

Question 15

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 it reaches and the total time

it takes for the projectile to hit the ground.

Solution

Step 1: Resolve the initial velocity into its x and y components. The initial

velocity v0can be resolved into its x-component v0xand y-component v0yusing

the following equations:

v0x=v0cos(θ)

v0y=v0sin(θ)

Substitute v0= 20 m/s and θ= 30◦:

v0x= 20 cos(30◦) = 20 ×√3

2= 10√3 m/s

v0y= 20 sin(30◦) = 20 ×1

2= 10 m/s

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

tile reaches the maximum height when its vertical velocity component becomes

zero. The time taken to reach the maximum height can be determined using

the following kinematic equation:

vyf =v0y−gt

Where vyf = 0 (at maximum height) and g= 9.81 m/s2. Substitute the known

values into the equation:

0 = 10 −9.81t

t=10

9.81 ≈1.02 s

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

be calculated using the following kinematic equation:

h=v0yt−1

2gt2

Substitute v0y= 10 m/s, t= 1.02 s, and g= 9.81 m/s2:

h= 10 ×1.02 −1

2×9.81 ×(1.02)2

h≈5.1−5.1≈0 m

Step 4: Calculate the total time of flight. The total time of flight is twice

the time taken to reach the maximum height, since the flight is symmetrical.

Therefore, the total time Tis:

T= 2t= 2 ×1.02

T≈2.04 s

Therefore, the maximum height reached by the projectile is 0 m and the

total time it takes to hit the ground is 2.04 s.

Question 16

Question

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

of 25 m/s. Calculate the total time the baseball is in the air and the maximum

height it reaches. Take g= 9.81 m/s2for the acceleration due to gravity.

Solution

Step 1: Resolve initial velocity into horizontal and vertical components: Let

v0= 25 m/s be the initial velocity and θ= 35◦be the launch angle. The initial

velocity components are:

v0x=v0cos θ

v0y=v0sin θ

Step 2: Calculate time to reach maximum height: The time to reach maxi-

mum height can be found by using the vertical component of the initial velocity

and acceleration due to gravity:

v0y=vy−gt

At maximum height, vy= 0:

0 = v0y−gt

t=v0y

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

baseball can be calculated using the equation for the vertical displacement at

maximum height:

y=v2

0y−2g∆y

At maximum height, vy= 0:

0 = v2

0y−2g∆y

∆y=v2

Step 4: Calculate total time in the air: The total time in the air is twice the

time to reach maximum height:

Total time = 2t

Substitute the given values to find the answers. - v0y= 25 ·sin 35◦-t=

25·sin 35◦

9.81 - ∆y=(25·sin 35◦)2

2·9.81 - Total time = 2 ·25·sin 35◦

9.81

Question 17

Question

A baseball is hit at an angle of 40◦above the horizontal with an initial speed of

25 m/s. Calculate the maximum height the baseball reaches during its flight.

Solution

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

components.

vix=vi·cos(θ)

viy=vi·sin(θ)

Step 2: Substitute the given values into the equations.

vix= 25 m/s ·cos(40◦)≈19.12 m/s

viy= 25 m/s ·sin(40◦)≈16.06 m/s

Step 3: Calculate the time taken for the baseball to reach its maximum

height using the vertical component of the initial velocity.

vf=viy+agravity ·t

Since the baseball reaches its maximum height, the final velocity is 0 m/s. So,

0 m/s = 16.06 m/s −9.81 m/s2·t

Step 4: Solve for the time t.

t=16.06 m/s

9.81 m/s2≈1.64 s

Step 5: Calculate the maximum height by using the equation for vertical

displacement when the final velocity is zero.

ymax =viy·t−1

2agravity ·t2

ymax = 16.06 m/s ·1.64 s −1

2·9.81 m/s2·(1.64 s)2

ymax ≈13.2 m

Therefore, the maximum height the baseball reaches during its flight is ap-

proximately 13.2 meters.

Question 18

Question

A particle moves in the xy plane with a position vector r = 3t2ˆ

i+(5−2t)ˆ

j, where

r is in meters and tis in seconds. Find the particle’s velocity and acceleration

as functions of t.

Solution

Step 1: To find the particle’s velocity, we differentiate the position vector r with

respect to time t.

Velocity v =dr

dt(3t2ˆ

i+ (5 −2t)ˆ

= 6tˆ

i−2ˆ

Step 2: To find the particle’s acceleration, we differentiate the velocity vector

v with respect to time t.

Acceleration a =dv

dt(6tˆ

i−2ˆ

= 6ˆ

Therefore, the particle’s velocity is v = 6tˆ

i−2ˆ

jm/s and the acceleration is

a = 6ˆ

im/s2as functions of time t.

Question 19

Question

A particle moves in the x-yplane with an acceleration given by a = (3t)ˆ

i−(4)ˆ

where ˆ

iand ˆ

jare unit vectors in the xand ydirections, respectively. If at t= 0

the particle is at the origin with a velocity of v0= (5 m/s)ˆ

i+ (2 m/s)ˆ

j, find the

velocity vector as a function of time.

Solution

Step 1: Integrate the acceleration vector a to get the velocity vector v.

Given: a = (3t)ˆ

i−(4)ˆ

Integrating each component with respect to time: Raxdt =R3t dt and

Raydt =R−4dt

We get: vx=R3t dt =1

2(3t2) = 3

2t2+Cx

vy=R−4dt =−4t+Cy

Step 2: Apply the initial conditions to find the constants Cxand Cy.

Given initial velocity v0= (5 m/s)ˆ

i+(2 m/s)ˆ

j, which means at t= 0: vx(0) =

5 and vy(0) = 2

Using these initial conditions: vx(0) = 3

2(0)2+Cx=Cx= 5 vy(0) =

−4(0) + Cy=Cy= 2

So, vx=3

2t2+ 5 and vy=−4t+ 2

Therefore, the velocity vector as a function of time is: v =3

2t2+ 5ˆ

(−4t+ 2)ˆ

Question 20

Question

A particle moves along a curved path defined by the parametric equations x(t) =

3t2and y(t)=4t3. Find the magnitude of the acceleration of the particle at

t= 2 seconds.

Solution

Step 1: Find the velocity vector v of the particle at t= 2 seconds by taking the

derivative of the position vector r(t) = x(t)ˆ

i+y(t)ˆ

jwith respect to time:

v =dr

dt =d

dt(3t2ˆ

i+ 4t3ˆ

j)=6tˆ

i+ 12t2ˆ

Now, evaluate v at t= 2 seconds:

v(2) = 6(2)ˆ

i+ 12(2)2ˆ

j= 12ˆ

i+ 48ˆ

Step 2: Find the acceleration vector a of the particle at t= 2 seconds by

taking the derivative of the velocity vector v with respect to time:

a =dv

dt =d

dt(6tˆ

i+ 12t2ˆ

j) = 6ˆ

i+ 24tˆ

Now, evaluate a at t= 2 seconds:

a(2) = 6ˆ

i+ 24(2)ˆ

j= 6ˆ

i+ 48ˆ

Step 3: Calculate the magnitude of the acceleration vector a at t= 2 seconds:

|a(2)|=p(6)2+ (48)2=√36 + 2304 = √2340 ≈48.38 m/s2

Therefore, the magnitude of the acceleration of the particle at t= 2 seconds

is approximately 48.38 m/s2.

Question 21

Question

A particle moves in the xy plane with an acceleration given by a = (4t2ˆ

i−

6tˆ

j) m/s2. If the particle starts from rest at the origin, find its velocity vector

as a function of time.

Solution

Step 1: To find the velocity vector as a function of time, we need to integrate the

acceleration function with respect to time. Step 2: Integrating the xcomponent

of acceleration, ax= 4t2m/s2, with respect to time, we get the xcomponent of

velocity:

Zaxdt =Z4t2dt =4

3t3+C1

where C1is the constant of integration. The xcomponent of velocity is vx=

3t3+C1. Step 3: Integrating the ycomponent of acceleration, ay=−6tm/s2,

with respect to time, we get the ycomponent of velocity:

Zaydt =Z−6t dt =−3t2+C2

where C2is the constant of integration. The ycomponent of velocity is vy=

−3t2+C2. Step 4: The velocity vector is given by v =vxˆ

i+vyˆ

j. Substituting

in the expressions for vxand vy, we have the velocity vector as a function of

time:

v =4

3t3+C1ˆ

i+−3t2+C2ˆ

where C1and C2are constants of integration.

Question 22

Question

A soccer player kicks the ball from the ground at an angle of 40◦above the

horizontal with an initial speed of 20 m/s. Find the maximum height the ball

reaches and the total time it is in the air before hitting the ground. Neglect air

resistance.

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 component

v0xand vertical component v0y:

v0x=v0cos(40◦)

v0y=v0sin(40◦)

Step 2: Find the time the ball takes to reach the maximum height. At

the maximum height, the vertical component of the velocity is 0 m/s. Using

the equation of motion vy=v0y−gt, where v0yis the initial vertical velocity,

g= 9.8 m/s2is the acceleration due to gravity, and tis the time taken to reach

the maximum height:

0 = v0y−gt

t=v0y

Step 3: Calculate the maximum height reached by the ball. Using the equa-

tion of motion y=y0+v0yt−1

2gt2, where y0= 0 m is the initial vertical

position, v0yis the initial vertical velocity, g= 9.8 m/s2is the acceleration due

to gravity, and tis the time taken to reach the maximum height:

ymax =v0y×v0y

g−1

2gv0y

g2

Step 4: Find the total time the ball is in the air. The total time the ball

is in the air is twice the time it takes to reach the maximum height since the

vertical motion is symmetrical.

Total time = 2 ×v0y

Question 23

Question

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

horizontal. Determine the maximum height the ball reaches and the total time

of flight.

Solution

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

Let the initial velocity vi= 20 m/s, the angle above the horizontal θ= 30◦. The

horizontal component vix=vicos θand the vertical component viy=visin θ.

vix= 20 m/s cos 30◦= 20 m/s ×√3

2= 10√3 m/s

viy= 20 m/s sin 30◦= 20 m/s ×1

2= 10 m/s

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

The time tit takes for the ball to reach the maximum height can be found using

the vertical motion equation y=viyt+1

2ayt2. Since the ball reaches maximum

height when vy= 0, we have 0 = viy−gtmax →tmax =viy

Substitute viy= 10 m/s and g= 9.8 m/s2:

tmax =10

9.8≈1.02 s

Step 3: Calculate the maximum height the ball reaches.

The maximum height hmax can be calculated using the vertical motion equation

y=v2

iy−2ayy. Since the final vertical velocity vy= 0 at the maximum height,

we have 0 = v2

iy−2ghmax.

Solve for hmax:

hmax =v2

2g=(10 m/s)2

2×9.8 m/s2≈5.10 m

Step 4: Determine the total time of flight.

Since the ball is thrown symmetrically, the total time of flight Tis twice the

time to reach maximum height, T= 2 ×tmax.

Substitute tmax = 1.02 s into the equation:

T= 2 ×1.02 s = 2.04 s

Therefore, the maximum height the ball reaches is approximately 5.10 m and

the total time of flight is 2.04 s.

Question 24

Question

A particle moves along the following path in the xy plane: it starts at the origin,

moves to the point (3 m,4 m), then to the point (−2 m,1 m), and finally ends

at the point (−1 m,−5 m). Calculate the total distance traveled by the particle.

Solution

Step 1: Calculate the distance between the initial and final points.

The distance between two points (x1, y1) and (x2, y2) in the xy plane is given

by the formula:

D=p(x2−x1)2+ (y2−y1)2

Using this formula, we can calculate the distances between the sequential points:

- From (0,0) to (3,4):

D1=p(3 −0)2+ (4 −0)2=p32+ 42= 5 m

- From (3,4) to (−2,1):

D2=p(−2−3)2+ (1 −4)2=p(−5)2+ (−3)2=√25 + 9 = √34 m

- From (−2,1) to (−1,−5):

D3=p(−1 + 2)2+ (−5−1)2=√1 + 36 = √37 m

Step 2: Calculate the total distance traveled.

The total distance is the sum of the individual distances:

Total distance = D1+D2+D3= 5 m+√34 m+√37 m ≈5 m+5.83 m+6.08 m ≈16.91 m

Therefore, the total distance traveled by the particle is approximately 16.91

meters.

Question 25

Question

A baseball player hits a ball at an angle of 35◦above the horizontal. The ball

leaves the bat with a speed of 30 m/s and an initial height of 1.5 m above the

ground. How far does the ball travel horizontally before hitting the ground?

Solution

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

Step 2: Find the time of flight of the ball. Step 3: Calculate the horizontal

distance the ball travels before hitting the ground.

Step 1: Resolve the initial velocity: The initial velocity of the ball can be

resolved into horizontal (V0x) and vertical (V0y) components using trigonometry:

V0x=V0cos θand V0y=V0sin θ

Substitute V0= 30 m/s and θ= 35◦into the above equations:

V0x= 30 cos 35◦≈24.65 m/s

V0y= 30 sin 35◦≈17.12 m/s

Step 2: Find the time of flight: The time of flight (t) can be found using

the equation for vertical motion:

y=V0yt+1

2gt2

where yis the initial height (1.5 m) and gis the acceleration due to gravity

(-9.81 m/s2). Since the ball lands back on the ground, the final vertical position

is 0. Solving for t:

0 = 17.12t+1

2(−9.81)t2

−4.905t2+ 17.12t= 0

t(17.12 −4.905t)=0

t= 0 or t=17.12

4.905 ≈3.49 s

Step 3: Calculate the horizontal distance: The horizontal distance the ball

travels can be calculated using the equation for horizontal motion:

x=V0xt

Substitute V0x= 24.65 m/s and t= 3.49 s into the above equation:

x= 24.65 ×3.49 ≈86.00 m

Therefore, the ball travels approximately 86.00 meters horizontally before hit-

ting the ground.

Question 26

Question

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

horizontal. Calculate the maximum height reached by the ball. Assume the

acceleration due to gravity is 9.8 m/s2.

Solution

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

The initial vertical component of velocity is: viy=visin θ= 20 m/s ×sin(30◦)

Step 2: Calculate the initial vertical component of velocity. viy= 20 m/s ×

2= 10 m/s

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

height. The equation for vertical motion is: y=yi+viyt−1

2gt2

At the maximum height, the final vertical velocity is zero. Therefore, vfy =

Step 4: Find the time taken to reach the maximum height. 0 = viy −gt

t=viy

Step 5: Substitute the values to find the maximum height. y= 0+(10 m/s)×

10 m/s

9.8 m/s2−1

2×9.8 m/s2×10 m/s

9.8 m/s22

Step 6: Calculate the maximum height. y= 0 + 10 m −1

2×9.8 m/s2×10

9.8

y= 10 m −5 m = 5 m

Therefore, the maximum height reached by the ball is 5 m.

Question 27

Question

A baseball player hits a ball with an initial velocity of 40 m/s at an angle of 30

degrees above the horizontal. Determine the maximum height the ball reaches

during its flight.

Solution

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

The horizontal component of the velocity is given by v0x=v0·cos(θ) and the

vertical component is given by v0y=v0·sin(θ). Given that v0= 40 m/s and

θ= 30◦, we have: v0x= 40·cos(30◦) = 40·√3

2= 20√3 m/s v0y= 40 ·sin(30◦) =

40 ·1

2= 20 m/s

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

The time it takes to reach the maximum height can be found using the vertical

component of the velocity and the acceleration due to gravity. The formula is

vf=vi+at, where vf= 0 at the maximum height. We have: 0 = v0y−gt

Solving for t, we get: t=v0y

g=20

9.8≈2.04 s

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

using the vertical motion equation y=v0yt−1

2gt2. Substitute the values we

obtained: y= 20 ·2.04 −1

2·9.8·(2.04)2Final calculation gives: y≈20.4 m

Therefore, the maximum height the ball reaches is approximately 20.4 me-

ters.

Question 28

Question

A hockey puck is initially at rest on the ice. It is then struck by a hockey

stick, giving it an initial velocity of 10 m/s at an angle of 30 degrees above the

horizontal. The coefficient of kinetic friction between the puck and the ice is

0.1. How far does the puck travel before coming to rest?

Solution

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

The initial velocity of the puck is 10 m/s at an angle of 30 degrees above

the horizontal. Let vixrepresent the initial velocity in the x-direction and viy

represent the initial velocity in the y-direction.

vix= 10 cos(30◦) = 8.66 m/s

viy= 10 sin(30◦) = 5 m/s

Step 2: Determine the acceleration of the puck in the x-direction. Since the

puck is on ice, the frictional force fk=µk·Nwill act against the motion of

the puck in the x-direction. The normal force Ncan be calculated using the

gravitational force.

N=mg

N= 0.1·9.8 m/s2= 0.98 N

fk= 0.1·0.98 = 0.098 N

ax=−fk

m=−0.098

1=−0.098 m/s2

Step 3: Determine the time for the puck to stop in the x-direction. The final

velocity in the x-direction is 0 m/s since the puck eventually comes to a stop.

vfx=vix+axt

0=8.66 −0.098t

t=8.66

0.098 = 88.3 s

Step 4: Calculate the distance traveled by the puck before coming to rest.

The distance traveled in the x-direction can be calculated using the equation:

dx=vixt+1

2axt2

dx= 8.66(88.3) + 0.5(−0.098)(88.3)2

dx= 763.7 m

Therefore, the puck travels 763.7 meters before coming to a stop.

Question 29

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 reached by the projectile.

Solution

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

Given: Initial speed, v0= 30 m/s Launch angle, θ= 60◦

The horizontal component of the initial velocity is:

v0x=v0cos θ

v0x= 30 cos 60◦

v0x= 30 ×1

v0x= 15 m/s

The vertical component of the initial velocity is:

v0y=v0sin θ

v0y= 30 sin 60◦

v0y= 30 ×√3

v0y= 15√3 m/s

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

to reach maximum height. The equation for vertical motion is:

y=v0yt−1

2gt2

At maximum height, the vertical velocity is zero. So, vy= 0 Hence, we can use

the equation:

0 = v0y−gt

Solving for t:

15√3=9.8t

t≈15√3

9.8

t≈2.73 sec

Step 3: Find the height attained by the projectile. Using the displacement

formula in the vertical direction:

y=v0yt−1

2gt2

Substitute the known values:

y= 15√3×2.73 −1

2×9.8×(2.73)2

y≈30.88 m

Therefore, the maximum height reached by the projectile is approximately

30.88 meters.

Question 30

Question

A cannonball is fired with an initial speed of 100 m/s at an angle of 30◦above

the horizontal. Determine the maximum height the cannonball reaches and the

total time it is in the air.

Solution

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

Given that the initial speed is 100 m/s and the angle of projection is 30◦, the

horizontal component of velocity (v0x) and the vertical component of velocity

(v0y) can be calculated as follows:

v0x=v0cos(30◦)

v0y=v0sin(30◦)

Step 2: Calculate the time of flight. Using the kinematic equation y=

v0yt+1

2at2, we can find the time it takes for the cannonball to reach its max-

imum height. At the maximum height, the vertical component of velocity is 0.

Therefore:

0 = v0y−gt

t=v0y

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

using the kinematic equation y=v0yt−1

2gt2. Substituting the time of flight

calculated in Step 2:

y=v0y·v0y

g−1

2g·v0y

g2

Step 4: Calculate the total time in the air. As the total time in the air is

twice the time of flight, the total time in the air is:

Total time = 2t

Question 31

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 reached by the projectile.

Solution

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

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

components as follows: The horizontal component: v0x=v0cos(θ) Given that

v0= 30 m/s and θ= 60◦, we have: v0x= 30 cos(60◦) = 30 ×1

2= 15 m/s

The vertical component: v0y=v0sin(θ)v0y= 30 sin(60◦) = 30×√3

2= 15√3

m/s

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

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

initial velocity and acceleration due to gravity. The equation relating initial

velocity, acceleration, time, and displacement in the vertical direction is: vy=

v0y−gt, where vy= 0 at maximum height. Solving for tgives: t=v0y

15√3

9.81 ≈1.54 s

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

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

component of the initial velocity, time to reach maximum height, and accelera-

tion due to gravity. The equation for vertical displacement is: h=v0yt−1

2gt2

Substitute v0y= 15√3 m/s, t≈1.54 s, and g= 9.81 m/s2into the above equa-

tion to find the maximum height: h= 15√3×1.54 −1

2×9.81 ×(1.54)2≈23.83

Therefore, the maximum height reached by the projectile is approximately

23.83 meters.

Question 32

Question

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

an angle of 60 degrees above the horizontal. Find the projectile’s total time of

flight and maximum height reached. Take the acceleration due to gravity as

−9.81 m/s2.

Solution

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



Let the initial velocity v0= 30 m/s and the angle of elevation θ= 60◦.



The horizontal component of the initial velocity v0x=v0cos θ.



The vertical component of the initial velocity v0y=v0sin θ.



Substituting v0= 30 m/s and θ= 60◦into the above formulas: v0x=

30 cos 60◦and v0y= 30 sin 60◦.

Step 2: Calculate the time of flight using the vertical component of motion.



The time of flight in projectile motion can be determined by considering

the vertical component.



The formula vf=v0+at can be used to find the time taken to reach

maximum height.



At the peak of the projectile’s trajectory, the vertical component of the

final velocity vfy = 0.



Substituting vfy = 0, v0y= 30 sin 60◦, and a=−9.81 m/s2into the

formula: 0 = 30 sin 60◦−9.81t. Solve for t.

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



The maximum height can be calculated using the vertical motion equation

y=v0yt+1

2at2.



The time tfound in Step 2 can be used to calculate the maximum height

ymax.



Substituting v0y= 30 sin 60◦,tfrom Step 2, and a=−9.81 m/s2into the

above equation.

Question 33

Question

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

of 20 m/s from the edge of a cliff that is 50 m high. How far from the base of

the cliff does the projectile strike the ground?

Solution

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

The initial velocity (v0) of the projectile can be resolved into its horizontal

component (v0x) and vertical component (v0y) as follows:

v0x=v0cos θ

v0y=v0sin θ

where v0= 20 m/s and θ= 30◦.

Step 2: Calculate the time taken for the projectile to hit the ground. The

time taken for the projectile to hit the ground can be found using the vertical

motion equation:

y=v0yt+1

2at2

where y= -50 m (height of the cliff), v0y=v0sin θ,a= -9.8 m/s2(acceleration

due to gravity), and tis the time of flight. Solving for tgives:

−50 = (20 sin 30◦)t−1

2(9.8)t2

Step 3: Find the horizontal distance traveled by the projectile. The horizon-

tal distance traveled by the projectile can be determined using the horizontal

motion equation:

x=v0xt

where v0x=v0cos θ. Substituting the known values into the equation gives:

x= (20 cos 30◦)t

Step 4: Calculate the horizontal distance from the base of the cliff. Finally,

substitute the value of tfrom Step 2 into the equation from Step 3 to find the

horizontal distance:

x= (20 cos 30◦)t

Question 34

Question

A projectile is launched 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

projectile.

Solution

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

The initial velocity can be resolved into two components: - The horizontal

component vix=vicos(θ), - The vertical component viy=visin(θ),

where viis the initial velocity (20 m/s) and θis the launch angle (30◦).

Calculating the components: vix= 20 cos(30◦) = 20 ·√3

2= 10√3 m/s, viy=

20 sin(30◦) = 20 ·1

2= 10 m/s.

Step 2: Determine the time it takes for the projectile to reach the maximum

height. Since the projectile reaches maximum height when the vertical compo-

nent of velocity is zero, we can use the equation vf=vi−gt, where vfis the

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

Substitute into the equation: 0 = 10 −9.8t, Solving for tgives: t=10

9.8= 1.02

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

found using the equation: h=viyt−1

2gt2. Substitute the known values: h=

10 ·1.02 −1

2·9.8·(1.02)2= 10.2−5·1.02 = 10.2−5.1 = 5.1 m.

Therefore, the maximum height reached by the projectile is 5.1 meters.

Question 35

Question

A stone is thrown at an angle of 30◦above the horizontal with an initial speed

of 20 m/s from the edge of a cliff 80 m high. Find the total time the stone is in

the air.

Solution

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

components. The horizontal component of the initial velocity V0xis given by:

V0x=V0cos(θ)

where V0is the initial speed and θis the angle of projection. Substitute V0=

20 m/s and θ= 30◦into the equation:

V0x= 20 cos(30◦)

V0x= 20 ·√3

V0x= 10√3 m/s

The vertical component of the initial velocity V0yis given by:

V0y=V0sin(θ)

Substitute V0= 20 m/s and θ= 30◦into the equation:

V0y= 20 sin(30◦)

V0y= 20 ·1

V0y= 10 m/s

Step 2: Determine the time taken for the stone to reach the peak of its

trajectory. Use the vertical motion equation: y=V0yt−1

2gt2. At the peak

of the trajectory, the stone reaches a height of 80 m, so y= 80 m. Substitute

V0y= 10 m/s and g= 9.81 m/s2into the equation:

80 = 10t−1

2·9.81 ·t2

Step 3: Solve for the time taken to reach the peak of the trajectory. Rear-

range the equation to get a quadratic equation in the form at2+bt +c= 0:

−4.905t2+ 10t−80 = 0

Using the quadratic formula t=−b±√b2−4ac

2awith a=−4.905, b= 10, and

c=−80:

t=−10 ±p102−4(−4.905)(−80)

2(−4.905)

t=−10 ±√100 + 1962

−9.81

t=−10 ±√2062

−9.81

t≈2.78 s

So, it takes approximately 2.78 seconds for the stone to reach the peak of

its trajectory.

PHYS 202 - GENERAL PHYSICS II Kinematics in Two Dimensions Question Bank Set 5

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