PHYS 202 - GENERAL PHYSICS II Kinematics in Two Dimensions Question Bank Set 3

PHYS 202 - GENERAL PHYSICS II -

Kinematics in Two Dimensions

Question Bank - Set 3

Liberty University

Question 1

Question

A particle moves along a curved path in the xy-plane with position given by

r(t) = (4t3−2t2)ˆ

i+ (3t2−5t)ˆ

j, where tis in seconds. Find the magnitude of

the particle’s acceleration at t= 2 s.

Solution

Step 1: To find the acceleration of the particle, we first need to calculate its

velocity and then differentiate the velocity to find the acceleration.

Step 2: The velocity of the particle is given by v(t) = dr

dt . Differentiating

r(t) with respect to t, we have:

v(t) = (12t2−4t)ˆ

i+ (6t−5)ˆ

Step 3: Now, to find the acceleration, we differentiate the velocity with

respect to t:

a(t) = dv

dt = (24t−4)ˆ

i+ 6ˆ

Step 4: To find the magnitude of the acceleration at t= 2 s, we substitute

t= 2 into the expression for a(t):

a(2) = (24(2) −4)ˆ

i+ 6ˆ

j= 44ˆ

i+ 6ˆ

Step 5: The magnitude of the acceleration is given by |a(2)|=p(44)2+ (6)2=

√1936 + 36 = √1972 ≈44.39 m/s2.

Therefore, the magnitude of the particle’s acceleration at t= 2 s is approx-

imately 44.39 m/s2.

Question 2

Question

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

horizontal with a velocity of 20 m/s. Calculate the maximum height the ball

reaches and the horizontal distance it travels before hitting the ground.

Solution

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

horizontal component of the initial velocity is given by v0x=v0cos(θ) where

v0= 20 m/s and θ= 30◦. The vertical component of the initial velocity is given

by v0y=v0sin(θ).

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

the vertical motion is influenced by gravity, we can use the kinematic equation

vy=v0y−gt where vy= 0 at the maximum height and g= 9.8 m/s2. Solving

for t, we get t=v0y

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

height can be calculated using the kinematic equation y=v0yt−1

2gt2. Substi-

tute v0yand tinto the equation to find the maximum height.

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

twice the time it takes to reach the maximum height, ttotal = 2t.

Step 5: Calculate the horizontal distance traveled. The horizontal distance

can be calculated using the equation x=v0xttotal. Substitute v0xand ttotal into

the equation to find the horizontal distance.

Question 3

Question

A baseball player hits a ball with an initial velocity of 25 m/s at an angle of

30◦above the horizontal. The ball lands on the ground 120 m away. Find 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 initial velocity is given by vix=vicos θ, where

vi= 25 m/s and θ= 30◦.

vix= 25 cos 30◦= 25 ×√3

2≈21.65 m/s

The vertical component of the initial velocity is given by viy=visin θ.

viy= 25 sin 30◦= 25 ×1

2= 12.5 m/s

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

the horizontal component of the velocity. Since the horizontal acceleration is

zero, the horizontal velocity remains constant. Using the equation s=vix·t,

where s= 120 m and vix= 21.65 m/s:

t=s

vix

=120

21.65 ≈5.54 s

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

height can be found using the vertical component of the velocity and the time

of flight. The equation for the maximum height reached is:

h=viy·t−1

2·g·t2

where g≈9.81 m/s2is the acceleration due to gravity. Plugging in the values:

h= 12.5×5.54 −1

2×9.81 ×(5.54)2

h≈35.71 m −153.09 m

h≈ −117.38 m

Since the negative value indicates the ball is below the starting point, the

maximum height reached by the ball during its flight is 117.38 m .

Question 4

Question

A football player kicks the ball with an initial velocity of 20 m/s at an angle of

45◦above the horizontal. Find the maximum height the ball reaches and the

total time the ball is in the air. Assume air resistance is negligible.

Solution

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

The initial velocity (v0) can be resolved into its horizontal component (v0x) and

vertical component (v0y) using trigonometric functions. Given v0= 20 m/s and

the angle θ= 45◦, we have:

v0x=v0cos θ

v0y=v0sin θ

Step 2: Calculate the time to reach maximum height. As the ball reaches

its maximum height, its vertical velocity component becomes zero (vy= 0). We

can use the kinematic equation:

vy=v0y−gt

Substitute vy= 0 and solve for t:

0 = v0sin θ−gt

t=v0sin θ

Step 3: Calculate the maximum height reached. The maximum height (h)

can be found using the kinematic equation:

h=v0yt−1

2gt2

Substitute the values of v0yand tinto the equation and solve for h:

h= (v0sin θ)v0sin θ

g−1

2gv0sin θ

g2

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

ball is in the air is twice the time calculated in Step 2, since the time to reach

maximum height is equal to the time to fall back down. Therefore, the total

time in the air is:

Total time = 2t= 2 v0sin θ

g

Question 5

Question

A baseball is hit with an initial velocity of 30 m/s at an angle of 60◦above the

horizontal. Calculate the maximum height the baseball reaches during its flight.

(Assume no air resistance)

Solution

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

The horizontal component vixis given by:

vix=vicos(θ)

where vi= 30 m/s and θ= 60◦.

vix= 30 cos(60◦) = 30 ×1

2= 15 m/s

The vertical component viyis given by:

viy=visin(θ)

viy= 30 sin(60◦) = 30 ×√3

2= 15√3 m/s

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

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

t=vf−vi

−g

where vi= 15√3 m/s (initial vertical velocity), vf= 0 (final vertical velocity at

maximum height), and g= 9.81 m/s2(acceleration due to gravity).

t=0−15√3

−9.81 ≈1.53 s

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

mum height can be found using the formula:

ymax =viyt−1

2gt2

Substitute viy= 15√3 m/s and t= 1.53 s into the equation:

ymax = 15√3×1.53 −1

2×9.81 ×(1.53)2≈33.9 m

Therefore, the maximum height the baseball reaches is approximately 33.9

meters.

Question 6

Question

A projectile is launched with an initial speed of 50 m/s at an angle of 30 degrees

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

during its flight.

Solution

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

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

follows: The horizontal component: vix=vi·cos(θ) The vertical component:

viy=vi·sin(θ) where vi= 50 m/s is the initial speed, θ= 30◦is the launch

angle.

Substitute the values: vix= 50 ·cos(30◦)viy= 50 ·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 found using the

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

which is −9.81 m/s2.

The vertical displacement yat maximum height is zero. The equation of

motion for vertical motion is: y=viy·t−1

2·g·t2where y= 0, viy= 50·sin(30◦),

g=−9.81 m/s2, and we need to solve for t.

Step 3: Find the maximum height. Once we have the time taken to reach the

maximum height, we can find the vertical distance traveled using the equation:

y=viy·t−1

2·g·t2.

Question 7

Question

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

above the horizontal. The ball lands on the ground 150 meters away. Calculate

the maximum height the ball reaches during its flight.

Solution

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

initial velocity of the ball can be resolved into horizontal and vertical components

as follows: Horizontal component: vix=vicos θ= 30 m/s ×cos 30◦≈25.98 m/s

Vertical component: viy=visin θ= 30 m/s ×sin 30◦= 15 m/s

Step 2: Use the vertical motion equation to find the maximum height. The

equation for vertical motion is: y=viyt−1

2gt2where yis the height, viyis the

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

taken to reach the maximum height.

At the maximum height, the vertical velocity becomes zero. Therefore, we

can find the time taken to reach the maximum height using: 0 = viy−gt Solving

for tgives: t=viy

g=15 m/s

9.8 m/s2≈1.53 s

Now, substitute t= 1.53 s into the vertical motion equation to find the

maximum height ymax:ymax =viyt−1

2gt2= 15 m/s ×1.53 s −1

2×9.8 m/s2×

(1.53 s)2≈11.52 m

Therefore, the maximum height the ball reaches during its flight is approxi-

mately 11.52 meters.

Question 8

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 and the total time of flight. Take g= 9.81 m/s2.

Solution

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

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

follows:

v0x=v0cos(θ)

v0y=v0sin(θ)

Given that v0= 20 m/s and θ= 30◦, we have:

v0x= 20 cos(30◦)≈17.32 m/s

v0y= 20 sin(30◦) = 10 m/s

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

using the vertical component of motion. The projectile reaches its maximum

height when the vertical component of the velocity is zero. Using the equation

for vertical velocity, we have:

vy=v0y−gt

At the maximum height, vy= 0, so:

0 = 10 −9.81t

Solving for t, we get:

t=10

9.81 ≈1.02 seconds

Since at the maximum height, the object spends half of the time of flight, the

total time of flight is given by:

Time of flight = 2t≈2(1.02) ≈2.04 seconds

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

be calculated using the vertical component of motion. We can use the kinematic

equation:

y=v0yt−1

2gt2

Substitute v0y,t, and ginto the equation:

y= 10(1.02) −1

2(9.81)(1.02)2≈5.1 meters

Therefore, the maximum height reached by the projectile is approximately

5.1 meters, and the total time of flight is approximately 2.04 seconds.

Question 9

Question

A ball is thrown with an initial velocity of 15 m/s at an angle of 60 degrees above

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

Solution

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

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

follows:

vix =vicos(θ) = 15 m/s ·cos(60◦)

viy =visin(θ) = 15 m/s ·sin(60◦)

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

vertical velocity at the maximum height is 0, we can use the equation vf=vi+at

for the vertical component to find the time taken to reach the maximum height.

0 = viy −gtmax

tmax =viy

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

using the vertical displacement equation:

∆y=viyt−1

2gt2

∆y=viytmax −1

2g(tmax)2

Substitute the values of viy and tmax:

∆y= 15 m/s ·sin(60◦)15 m/s ·sin(60◦)

9.81 −1

2·9.81 15 m/s ·sin(60◦)

9.81 2

∆y=. . .

(Calculate the final answer to find the maximum height)

Question 10

Question

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

30 degrees above the horizontal. At the same moment, a teammate standing 60

meters away from the player, starts running directly away from the player at a

constant speed of 5 m/s. How far away from the player does the ball land?

Solution

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

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

vix=vi·cos(θ) = 30 m/s ·cos(30◦) = 26 m/s

The vertical component of the initial velocity is given by:

viy=vi·sin(θ) = 30 m/s ·sin(30◦) = 15 m/s

Step 2: Determine the time of flight of the ball. The time of flight can

be found using the vertical motion equation y=viyt+1

2ayt2, where yis the

vertical displacement, viyis the initial vertical velocity, ayis the acceleration in

the y-direction (gravity), and tis the time of flight.

Since the ball lands at the same height it was thrown, y= 0. Therefore the

equation reduces to:

0 = 15t−1

2·9.8 m/s2·t2

9.8t2−15t= 0

t(9.8t−15) = 0

So the time of flight is t= 0 s (at launch) and t=15

9.8≈1.53 s.

Step 3: Determine the horizontal distance the ball travels during this time.

The horizontal distance is given by x=vix·t= 26 m/s ·1.53 s ≈39.78 m.

Step 4: Determine the distance the teammate covers during this time. The

distance the teammate covers is given by dteammate = speed ·t= 5 m/s ·1.53 s =

7.65 m.

Step 5: Determine the total horizontal distance from the player where the

ball lands. The total horizontal distance is the sum of the distance the ball

travels and the distance the teammate covers:

dtotal = 39.78 m + 7.65 m = 47.43 m

Therefore, the ball lands approximately 47.43 meters away from the player.

Question 11

Question

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

above the horizontal. Calculate the maximum height the ball reaches and the

total time the ball is in the air. (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 velocity in the x-direction (v0x) is given by

v0x=v0cos(θ)

v0x= 20 cos(30◦)

v0x≈17.32 m/s

The initial velocity in the y-direction (v0y) is given by

v0y=v0sin(θ)

v0y= 20 sin(30◦)

v0y= 10 m/s

Step 2: Calculate the maximum height (H) the ball reaches. Use the kine-

matic equation for vertical motion:

f=v2

i+ 2a∆y

where vf= 0 (at maximum height) and a=−9.8 m/s2(acceleration due to

gravity).

0 = (10 m/s)2+ 2(−9.8)∆y

0 = 100 −19.6∆y

∆y=100

19.6

∆y≈5.1 m

So, the maximum height the ball reaches is 5.1 meters.

Step 3: Calculate the total time of flight. Use the kinematic equation for

vertical motion:

∆y=v0yt+1

2at2

5.1 = 10t+1

2(−9.8)t2

5.1 = 10t−4.9t2

4.9t2−10t+ 5.1=0

Solving this quadratic equation gives two possible values for t.

t≈1.02 s

t≈1.04 s

Since the total time in the air is the sum of the times it takes to go up and come

down, the total time is approximately 2.06 seconds.

Question 12

Question

A golf ball is struck and launches at an angle of 40◦above the horizontal with an

initial speed of 50 m/s. Calculate the maximum height the ball reaches during

its flight. (Assume air resistance is negligible)

Solution

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

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

components as follows: Vix=Vicos θ= 50 cos 40◦≈38.36 m/s Viy=Visin θ=

50 sin 40◦≈32.11 m/s

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

taken for the golf ball to reach the maximum height can be found using the

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

0 (at maximum height) Using the kinematic equation Vf=Vi+at where a=

−9.8 m/s2: 0 = 32.11 −9.8t t =32.11

9.8≈3.28 s

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

imum height can be calculated using the vertical component of the velocity

at the maximum height and the time taken to reach that height: hmax =

Viyt+1

2(−9.8)t2hmax = 32.11 ×3.28 + 1

2(−9.8)(3.28)2≈52.47 m

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

proximately 52.47 m.

Question 13

Question

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

horizontal with an initial speed of 20 m/s. How far from the player does the

ball land?

Solution

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

The initial velocity (v0) can be resolved into horizontal (v0x) and vertical (v0y)

components: v0x=v0cos(θ) and v0y=v0sin(θ), where v0is 20 m/s and θis 30◦.

Plugging in the values: v0x= 20 cos(30◦)≈17.3 m/s and v0y= 20 sin(30◦)≈

10 m/s.

Step 2: Determine the time of flight. For the vertical motion, we can use

the equation y(t) = v0yt+1

2at2, where ais the acceleration due to gravity (-9.8

m/s

). The ball lands when y(t) = 0. Substitute the known values into the

equation: 0 = 10t−4.9t2. Solving for t, we get two possible solutions: t= 0

(initial time) or t= 2 seconds. Since the time cannot be zero, the time of flight

is 2 seconds.

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

travels (x) can be determined using the equation x=v0x·t. Substitute v0x=

17.3 m/s and t= 2 s into the equation: x= 17.3×2 = 34.6 m.

Therefore, the ball lands approximately 34.6 meters from the player.

Question 14

Question

A ball is launched at an angle of 45◦above the horizontal from the top of a hill

that slopes downward at an angle of 30◦. The magnitude of its velocity is 20

m/s. Calculate the time it takes for the ball to hit the ground.

Solution

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

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

components as follows:

vi,x =vicos α

vi,y =visin α

where viis the magnitude of the initial velocity (20 m/s) and αis the launch

angle (45◦). Substituting the given values into the above equations gives:

vi,x = 20 cos 45◦= 20 ·1

√2= 10√2 m/s

vi,y = 20 sin 45◦= 20 ·1

√2= 10√2 m/s

Step 2: Find the time of flight for the vertical motion. The time of flight

(tflight) for the vertical motion of the ball can be determined using the vertical

motion equation:

y=vi,yt+1

2gt2

Since the ball is launched from the top of the hill, its initial vertical position

(yi) is 0. The final vertical position (yf) is also 0 since the ball hits the ground.

Therefore, the vertical equation simplifies to:

0 = (10√2)t−1

2·9.81 ·t2

10√2t−4.905t2= 0

t(10√2−4.905t)=0

This equation has two solutions: t= 0 and t=10√2

4.905 . Since time cannot be

negative, the time of flight for the vertical motion is t=10√2

4.905 ≈2.03 seconds.

Step 3: Find the time of flight for the horizontal motion. The time of flight

(thill) for the horizontal motion can be determined using the horizontal motion

equation:

x=vi,xthill

The horizontal distance traveled by the ball (x) is determined by the horizontal

component of the ball’s velocity and the time it takes for the ball to hit the

ground. The horizontal distance can be calculated using trigonometry and is

given by:

x=hcot γ

where his the height of the hill and γis the slope angle of the hill (30◦).

Substituting the given values:

x=hcot 30◦=h·√3

Step 4: Combine the horizontal and vertical motions. Since the ball hits the

ground after the same amount of time in both horizontal and vertical motions,

we can equate thill to tflight:

vi,xthill =tflight

10√2thill =10√2

4.905

Solving for thill gives:

thill =1

4.905 ≈0.204 seconds

Therefore, the time it takes for the ball to hit the ground is approximately

0.204 seconds.

Question 15

Question

A particle moves in the xy plane with an acceleration given by a = (4t−2)ˆ

i+6ˆ

where tis in seconds and the particle’s initial velocity is v0= 2ˆ

i+ 3ˆ

jm/s. Find

the magnitude and direction of the particle’s velocity at t= 2 s.

Solution

Step 1: Find the particle’s velocity at time tby integrating the acceleration

function with respect to time.

Za dt =Z(4t−2)ˆ

i+ 6ˆ

j dt

= (2t2−2t)ˆ

i+ 6tˆ

j+

where 

Cis the constant of integration.

Step 2: Use the initial velocity to find the constant of integration 

v0= 2ˆ

i+ 3ˆ

j= (2(0)2−2(0))ˆ

i+ (6(0))ˆ

j+



C= 2ˆ

i+ 3ˆ

Step 3: Substitute the acceleration function and constant of integration into

the velocity function.

v = (2t2−2t)ˆ

i+ 6tˆ

j+ 2ˆ

i+ 3ˆ

Step 4: Determine the velocity at t= 2 s.

v(2) = (2(2)2−2(2))ˆ

i+ 6(2)ˆ

j+ 2ˆ

i+ 3ˆ

= 4ˆ

i+ 12ˆ

j+ 2ˆ

i+ 3ˆ

= 6ˆ

i+ 15ˆ

Step 5: Calculate the magnitude and direction of the particle’s velocity at

t= 2 s.

|v(2)|=p(6)2+ (15)2

|v(2)|=√36 + 225

|v(2)|=√261 m/s

The direction of the velocity can be found by calculating the angle using the

arctangent function.

θ= arctan 15

6

θ≈67.38◦

Therefore, at t= 2 s, the magnitude of the particle’s velocity is √261 m/s at

an angle of 67.38◦above the positive x-axis.

Question 16

Question

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

speed of 40 m/s. Determine the maximum height reached by the projectile.

Solution

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

vertical component is given by vi,y =visin θwhere viis the initial speed (40

m/s) and θis the launch angle (30◦).

Step 2: Calculate the initial vertical component:

vi,y = 40 sin(30◦)≈20 m/s

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

the maximum height. The equation is vf=vi+at, where vfis the final velocity

(0 m/s at the maximum height), viis the initial velocity (20 m/s upwards), a

is the acceleration due to gravity (−9.8 m/s2downward), and tis the time:

0 = 20 −9.8t

Step 4: Solve for the time to reach the maximum height:

t=20

9.8≈2.04 s

Step 5: Use the kinematic equation for vertical motion to find the maximum

height above the launch point. The equation is y=viyt +1

2at2, where yis the

height, viyis the initial vertical velocity, tis the time, and ais the acceleration

due to gravity:

y= 20 ×2.04 + 1

2×(−9.8) ×(2.04)2

Step 6: Calculate the maximum height:

y≈20.4 m

Therefore, the maximum height reached by the projectile is approximately

20.4 meters.

Question 17

Question

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

of 20 m/s. At the highest point of its trajectory, the ball explodes into two

fragments of equal mass. One fragment, initially at rest, falls vertically. The

acceleration of gravity is 9.81 m/s2. What is the speed of the other fragment

just after the explosion?

Solution

Step 1: Calculate the initial velocity components of the ball.



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

vertical components. The horizontal component is vix=vicos(θ), where

vi= 20 m/s and θ= 45◦. Thus, vix= 20 cos(45◦) = 20 ×√2

2= 10√2 m/s.



The vertical component is viy=visin(θ), where vi= 20 m/s and θ= 45◦.

Thus, viy= 20 sin(45◦) = 20 ×√2

2= 10√2 m/s.

Step 2: Calculate the time taken for the ball to reach the highest point of

its trajectory.



The time taken to reach the highest point can be found using the vertical

component of velocity. The formula is vf=vi+at, where vf= 0 at the

highest point and a=−9.81 m/s2. Thus, 0 = 10√2−9.81t. Solving for t,

we get t=10√2

9.81 ≈1.42 s.

Step 3: Calculate the height reached by the ball.



The height reached by the ball can be found using the vertical component

of motion. The formula is y=viyt+1

2at2, where viy= 10√2 m/s, a=

−9.81 m/s2, and t≈1.42 s. Thus, y= (10√2×1.42)+ 1

2×−9.81×(1.42)2≈

10 m.

Step 4: Calculate the velocity of the other fragment just after the explosion.



The velocity of the other fragment after the explosion can be found using

the conservation of momentum. Since the fragments have equal mass, the

vertical component of velocity of the other fragment will be 10√2 m/s

as the original ball. Therefore, the speed of the other fragment just

after the explosion is v=p(vix)2+ (viy)2=q(10√2)2+ (10√2)2=

√200 + 200 = √400 = 20 m/s.

Question 18

Question

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

20 m/s. If air resistance is negligible, how far from the release point does the

ball hit the ground? (Take the acceleration due to gravity as −9.8 m/s2)

Solution

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

components. The initial velocity v0can be resolved into its horizontal (v0x) and

vertical (v0y) components as:

v0x=v0cos(30◦)

v0y=v0sin(30◦)

Step 2: Determine the time tit takes for the ball to hit the ground. Since the

vertical motion is solely under the influence of gravity, we can use the equation

of motion:

y=v0yt+1

2gt2

where yis the vertical displacement and gis the acceleration due to gravity.

Since the ball hits the ground, y= 0. Substituting the values of v0yand g

into the equation, we get:

0 = v0sin(30◦)t−1

2gt2

Solving for t, we get:

t=2v0sin(30◦)

Step 3: Calculate the horizontal distance the ball travels. Since the horizon-

tal motion of the ball is at constant velocity, the distance traveled horizontally

is given by:

x=v0xt

Substitute the values of v0xand tinto the equation:

x=v0cos(30◦)×2v0sin(30◦)

Step 4: Simplify to find the final answer. Calculating the value of x, we get:

x=2v2

0sin(30◦) cos(30◦)

x=v2

0sin(60◦)

Substitute v0= 20 m/s and g= 9.8 m/s2into the equation:

x=202×sin(60◦)

9.8≈34.87 m

Therefore, the ball hits the ground approximately 34.87 meters away from

the release point.

Question 19

Question

A projectile is launched from ground level with an initial velocity of 100 m/s at

an angle of 30◦above the horizontal. Find the horizontal and vertical compo-

nents of its velocity 1 second after launch.

Solution

Let’s break down the initial velocity into its horizontal and vertical components.

The horizontal component (v0x) can be found using the equation v0x=v0cos(θ)

and the vertical component (v0y) can be found using v0y=v0sin(θ).

Step 1: Find the horizontal and vertical components of the initial velocity:

Given: v0= 100 m/s, θ= 30◦

v0x= 100 cos(30◦) = 100 ·√3

2= 50√3 m/s

v0y= 100 sin(30◦) = 100 ·1

2= 50 m/s

So, v0x= 50√3 m/s and v0y= 50 m/s.

Step 2: Find the horizontal and vertical components of the velocity 1 second

after launch: The horizontal component of the velocity remains constant, while

the vertical component changes due to gravity.

Given: t= 1 s, ay=−9.8 m/s2(acceleration due to gravity)

vx=v0x= 50√3 m/s (horizontal component is constant)

To find the vertical component after 1 second, use the equation: vy=v0y+

ayt.

vy= 50 −9.8·1 = 40.2 m/s

Therefore, 1 second after launch, the horizontal component of velocity is

50√3 m/s and the vertical component is 40.2 m/s.

Question 20

Question

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

with an initial speed of 40 m/s. Find the time it takes for the projectile to reach

its maximum height.

Solution

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

The initial velocity vi= 40 m/s can be resolved into its horizontal and vertical

components using trigonometric functions. The vertical component is viy=

40 sin(30◦) m/s and the horizontal component is vix= 40 cos(30◦) m/s.

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

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

equation of motion vf=vi+at where a=−9.81 m/s2is the acceleration due

to gravity, we have: 0 = viy+ (−9.81)tSolving for t, we get: t=viy

9.81

Substitute viy= 40 sin(30◦) m/s into the equation to find t.

Question 21

Question

A stone is thrown off a cliff with an initial velocity of 20 m/s at an angle of 30◦

above the horizontal. The cliff is 50 m high. Calculate the time it takes for the

stone to hit the ground and the horizontal distance it travels before hitting the

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

Solution

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

The initial velocity has a magnitude of 20 m/s and makes an angle of 30◦with

the horizontal. The horizontal component is given by 20 m/s×cos(30◦) and the

vertical component is given by 20 m/s ×sin(30◦). So, the horizontal component

is:

v0x= 20 m/s ×cos(30◦) = 17.32 m/s

And the vertical component is:

v0y= 20 m/s ×sin(30◦) = 10 m/s

Step 2: Calculate the time it takes to hit the ground. Let’s find the time it

takes for the stone to hit the ground from the vertical motion. The equation

for vertical motion is:

y=v0y×t+1

2×g×t2

Where: y=−50 m (assuming downward direction as negative) v0y= 10 m/s

g=−9.81 m/s2We want to find t. Substitute the values into the equation:

−50 = 10t−1

2×9.81 ×t2

−50 = 10t−4.905t2

Rearrange the equation to form a quadratic equation:

4.905t2−10t−50 = 0

Solve the quadratic equation to find t. The positive value of twill be the time

it takes to hit the ground.

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

the stone travels is given by:

x=v0x×t

Substitute the known values and the time tfound in the previous step to calcu-

late the horizontal distance.

Question 22

Question

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

horizontal. The initial velocity of the ball is 20 m/s. Calculate the time it takes

for the ball to reach the highest point of its trajectory.

Solution

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

initial velocity of the ball vi= 20 m/s makes an angle of 30◦with the horizontal.

Therefore, the initial horizontal component of velocity is

vix=vicos(30◦) = 20 m/s ×cos(30◦).

And the initial vertical component of velocity is

viy=visin(30◦) = 20 m/s ×sin(30◦).

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

point of the trajectory, the vertical component of the velocity becomes zero. We

can use the equation for vertical motion:

vfy=viy−gt,

where vfyis the final vertical velocity, viyis the initial vertical velocity, gis the

acceleration due to gravity, and tis the time taken. Since at the highest point

vfy= 0, we can solve for t:

0 = viy−gt =⇒t=viy

Step 3: Substitute the values and calculate. Substitute viy= 20 m/s ×

sin(30◦) and g= 9.8 m/s2into the equation:

t=20 ×sin(30◦)

9.8.

t=20 ×1

9.8.

t=10

9.8≈1.02 s.

Therefore, it takes approximately 1.02 seconds for the ball to reach the

highest point of its trajectory.

Question 23

Question

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

horizontal with an initial speed of 20 m/s. Determine the projectile’s maximum

height above the ground. Neglect air resistance and assume the acceleration

due to gravity is 9.8 m/s2.

Solution

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

The initial velocity v0can be broken down into its horizontal component v0x

and vertical component v0y:

v0x=v0cos(θ)

v0y=v0sin(θ)

where v0= 20 m/s and θ= 45◦.

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

reach the maximum height can be determined using the vertical component:

vy=v0y−gt

At maximum height, the vertical velocity is 0, so:

0 = v0sin(θ)−gt

Solving for t:

t=v0sin(θ)

Step 3: Calculate the maximum height h. The maximum height is achieved

when the vertical velocity becomes zero. We can calculate the height using the

formula:

h=v0yt−1

2gt2

Substitute in the values:

h= (v0sin(θ)) v0sin(θ)

g−1

2gv0sin(θ)

g2

Step 4: Calculate the maximum height.

h= (20 sin(45◦)) 20 sin(45◦)

9.8−1

2·9.820 sin(45◦)

9.82

h= 20 ·√2

2! 20 ·√2

9.8!−1

2·9.8 20 ·√2

9.8!2

h=10√2 10√2

9.8!−1

2·9.8 10√2

9.8!2

h=200

9.8−1

2·9.8200

9.82

h=200

9.8−1

2·9.8·200

9.82

h=200

9.8−200

9.8

h= 0

Therefore, the maximum height above the ground is 0 meters. This result

indicates that the projectile does not reach any height and immediately lands

back on the ground.

Question 24

Question

A particle moves in the xy plane. At t= 0, its position vector with respect to

the origin is r(0) = (4.0 m)ˆ

i+ (3.0 m)ˆ

j. The particle has velocity v = (−2.0ˆ

3.0ˆ

j) m/s and acceleration a = (1.0ˆ

i−2.0ˆ

j) m/s2. Determine the position vector

of the particle as a function of time t.

Solution

Step 1: We can find the position vector as a function of time by integrating the

velocity and acceleration vectors with respect to time. The velocity vector v is

given by:

v =dr

Given that v =−2.0ˆ

i+ 3.0ˆ

j, we integrate each component separately:

For the x-component:

vx=dx

dt =−2.0

Integrating with respect to t:

Zdx =Z−2.0dt

x=−2.0t+C1

For the y-component:

vy=dy

dt = 3.0

Integrating with respect to t:

Zdy =Z3.0dt

y= 3.0t+C2

Thus, the velocity components are given by:

vx=−2.0t+C1

vy= 3.0t+C2

Step 2: Now, we can find the position vector as a function of time by inte-

grating the velocity components with respect to time.

Integrating the x-component:

x=Zvxdt =Z(−2.0t+C1)dt

x=−t2+C1t+C3

Integrating the y-component:

y=Zvydt =Z(3.0t+C2)dt

y= 1.5t2+C2t+C4

Thus, the position vector r as a function of time tis:

r(t)=(−t2+C1t+C3)ˆ

i+ (1.5t2+C2t+C4)ˆ

Question 25

Question

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

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

projectile explodes into two fragments. One fragment continued to move verti-

cally upward with a speed of 10 m/s immediately after the explosion, while the

other fragment continued to move horizontally at the same initial speed of the

projectile. Find the time interval between the explosion and the highest point

of the projectile’s trajectory.

Solution

Step 1: Break the initial velocity of the projectile into horizontal and vertical

components. Let v0= 20 m/s be the initial speed of the projectile, and θ= 30◦

be the launch angle. The horizontal component of the initial velocity is given

by v0x=v0cos θand the vertical component by v0y=v0sin θ.

Step 2: Determine the time to reach the highest point of the trajectory. The

time to reach the highest point can be found using the vertical component of the

motion. The vertical velocity at the highest point is zero. Using the equation

vy=v0y−gt, where g= 9.81 m/s2is the acceleration due to gravity, we have:

0 = v0y−gt

t=v0y

g=20 sin 30◦

9.81

Step 3: Calculate the vertical displacement at the highest point. The vertical

displacement at the highest point can be found using the equation y=v0yt−

2gt2. Plugging in the values:

y= 20 sin 30◦·20 sin 30◦

9.81 −1

2·9.81 ·20 sin 30◦

9.81 2

Step 4: Calculate the time intervals for the two fragments after the explosion.

The vertical fragment moves vertically upward with a speed of 10 m/s. The time

taken to reach the highest point is given by 10 = 20 sin 30◦−9.81tup. Solving

for tup, we get:

tup =20 sin 30◦−10

9.81

The horizontal fragment continues to move horizontally at the initial speed

of the projectile. Thus, the time taken by the horizontal fragment to reach the

highest point is the same as the time taken by the projectile, which we found

in Step 2.

Therefore, the time interval between the explosion and the highest point of

the projectile’s trajectory is:

Time Interval = tup −20 sin 30◦

9.81

Question 26

Question

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

above the horizontal. The initial velocity of the ball is 20 m/s. How high above

the ground does the ball go?

Solution

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

initial velocity of the ball can be resolved into horizontal and vertical components

using trigonometric identities. The vertical component can be found using vi,y =

visin(θ) and the horizontal component can be found using vi,x =vicos(θ).

Given: Initial velocity vi= 20 m/s Launch angle θ= 30◦

Calculations: vi,y = 20 m/s ×sin(30◦)≈10 m/s vi,x = 20 m/s ×cos(30◦)≈

17.32 m/s

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

reach the maximum height can be calculated using the vertical component of

initial velocity and the acceleration due to gravity. The vertical component of

acceleration is ay=−9.81 m/s2as gravity acts downward.

Using the formula: vf=vi+a×twhere vf= 0 at maximum height.

Substitute known values to find time:

0 = vi,y +ay×t0 = 10 −9.81 ×t t =10

9.81 ≈1.02 s

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

using the vertical displacement formula: y=vi,y ×t+1

2ay×t2.

Substitute known values to find the maximum height:

y= 10×1.02+ 1

2×(−9.81)×(1.02)2y= 10.2−5×1.0404 y≈10.2−5.202 ≈

4.998 m

Answer: The ball reaches a maximum height of approximately 4.998 meters

above the ground.

Question 27

Question

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

the horizontal.

1. What are the horizontal and vertical components of the initial velocity?

2. How long is the football in the air?

3. What is the maximum height reached by the football?

Solution

1. To find the horizontal and vertical components of the initial velocity, we can

use trigonometric identities. Let v0= 20 m/s be the magnitude of the initial

velocity.

Horizontal component = v0cos(30◦)

= 20 m/s ·cos(30◦)

= 20 m/s ·√3

= 10√3 m/s

Vertical component = v0sin(30◦)

= 20 m/s ·sin(30◦)

= 20 m/s ·1

= 10 m/s

2. The total time the football is in the air can be found by analyzing

the motion of the football vertically. The equation for vertical motion is y=

v0yt+1

2ayt2, where v0yis the vertical component of the initial velocity, ay=

−9.81 m/s2is the acceleration due to gravity, and yis the vertical displacement.

The football reaches its maximum height when its vertical velocity is zero. The

time taken to reach this point can be found using the equation vf y =v0y+ayt.

At maximum height: vfy = 0 ⇒0 = 10 −9.81t⇒t=10

9.81 ≈1.02 s

The total time in the air is twice this time, so the football is in the air for

2·1.02 s = 2.04 s.

3. The football’s maximum height can be found using the equation v2

fy =

0y+ 2ay∆y, where vf y = 0 m/s, v0y= 10 m/s, ay=−9.81 m/s2, and ∆yis the

maximum height.

02= (10)2+ 2(−9.81)∆y

∆y=(10)2

2·9.81

=100

19.62

≈5.10 m

Therefore, the maximum height reached by the football is approximately

5.10 meters.

Question 28

Question

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

angle of 60◦above the horizontal. Find the time it takes for the projectile to

reach its maximum height.

Solution

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

The initial velocity v0can be resolved into horizontal and vertical components

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

v0= 20 m/s and θ= 60◦. Therefore, v0x= 20 m/s ·cos(60◦). Calculating the

value gives v0x= 10 m/s.

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

θ= 60◦. Therefore, v0y= 20 m/s ·sin(60◦). Calculating the value gives v0y=

17.32 m/s.

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

the vertical component of velocity becomes zero. We can use the kinematic

equation for vertical motion: vf=vi+at where vf= 0 (final velocity), vi=v0y,

a=−9.81 m/s2(acceleration due to gravity), and we need to find t.

Substitute in the values: 0 = 17.32 m/s −9.81 m/s2·t. Solving for t, we get

t=17.32 m/s

9.81 m/s2. Calculating the value gives t≈1.77 s.

Therefore, it takes approximately 1.77 seconds for the projectile to reach its

maximum height.

Question 29

Question

A quarterback throws a football with an initial velocity of 25 m/s at an angle

of 35◦above the horizontal. The football is caught by the wide receiver 45 m

downfield. Find the height of the point from which the quarterback throws the

football.

Solution

Let’s break the initial velocity of the football into horizontal and vertical compo-

nents. The horizontal component is 25 m/s·cos(35◦) and the vertical component

is 25 m/s ·sin(35◦).

Step 1: Find the time taken for the football to travel 45 m downfield in

the horizontal direction. The horizontal distance traveled is given by d=vx·t,

where vx= 25 m/s ·cos(35◦). Solving for t:

t=d

=45 m

25 m/s ·cos(35◦)

Step 2: Find the height of the point from which the quarterback threw

the football. The height hat time tis given by h=viy·t−1

2·g·t2, where

viy= 25 m/s ·sin(35◦) and g= 9.81 m/s2. Substitute the values of viy,g, and t

into the formula to find h. Remember that at the time the football reaches the

receiver, the vertical position is equal to zero (h= 0), so his the height you are

looking for.

Question 30

Question

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

The projectile lands on the top of a 20 m high vertical wall that is a horizontal

distance of 40 m away from the launch point. Find the initial speed of the

projectile.

Solution

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

Let vibe the initial speed of the projectile. The initial velocity can be resolved

into two components: the horizontal component vix and the vertical component

viy. Since the angle of launch is 30◦, we have:

vix =vicos(30◦)

viy =visin(30◦)

Step 2: Calculate the time of flight. Using the vertical motion equation

y=viyt+1

2at2, where yis the vertical distance (20 m), viy is the initial vertical

component of velocity, ais the acceleration due to gravity, and tis the time of

flight, we have:

20 = visin(30◦)t−1

2gt2

Solving for t, we get:

t=2visin(30◦)

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

motion equation x=vixt, where xis the horizontal distance (40 m), vix is the

initial horizontal component of velocity, and tis the time of flight calculated in

Step 2, we have:

40 = vicos(30◦)2visin(30◦)

g

Step 4: Solve for the initial speed. Solving the equation in Step 3 for vigives

us:

vi=s40g

sin(60◦)

vi≈25.81 m/s

Therefore, the initial speed of the projectile is approximately 25.81 m/s.

Question 31

Question

A ball is thrown at an angle of 45◦above the horizontal with an initial speed of

20 m/s. Calculate the maximum height reached by the ball.

Solution

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

The initial velocity v0can be resolved into horizontal (v0x) and vertical (v0y)

components as follows:

v0x=v0cos(45◦) = 20 cos(45◦)≈14.14 m/s

v0y=v0sin(45◦) = 20 sin(45◦)≈14.14 m/s

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

taken for the vertical component of the velocity to become zero at the maximum

height is given by:

vy=v0y−gt

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

gravity, and tis the time taken. Setting vy= 0, we have:

0 = 14.14 −9.8t

t=14.14

9.8≈1.44 s

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

height hreached by the ball can be calculated using the equation for vertical

motion:

h=v0yt−1

2gt2

Substitute the values we found:

h= 14.14 ×1.44 −1

2×9.8×(1.44)2

h≈20.344 −10.475 ≈9.87 m

Therefore, the maximum height reached by the ball is approximately 9.87

meters.

Question 32

Question

A projectile is launched off of a cliff with an initial speed of 30 m/s at an angle

of 60 degrees above the horizontal. The cliff is 50 meters high and the projectile

lands 70 meters away from the base of the cliff. Calculate the time it takes for

the projectile to reach the ground.

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 using trigonometry. The horizontal component (vix) can be found

by vix =vicos(θ), and the vertical component (viy) can be found by viy =

visin(θ). Given: vi= 30 m/s, θ= 60◦.

Calculating the horizontal and vertical components: vix = 30 m/s·cos(60◦) =

15 m/s, viy = 30 m/s ·sin(60◦) = 25 m/s.

Step 2: Determine the time for the projectile to reach the ground. In the

vertical direction, the projectile experiences constant acceleration due to gravity.

The equation to determine the time it takes for the projectile to reach the

ground is y=viyt+1

2at2, where yis the vertical distance traveled (50 m in this

case), viy is the initial vertical velocity (25 m/s), ais the acceleration due to

gravity (−9.81 m/s2), and tis the time. Substitute the values into the equation:

50 m = 25 m/s ·t+1

2·(−9.81 m/s2)·t2.

Solving the equation for t:−4.905t2+ 25t−50 = 0.

Using the quadratic formula t=−b±√b2−4ac

2a, where a=−4.905, b= 25,

and c=−50.

t=−25±√252−4(−4.905)(−50)

2(−4.905) .

t≈5.10 s or −2.94 s.

Since time cannot be negative, the time it takes for the projectile to reach

the ground is approximately 5.10 s .

Question 33

Question

A particle moves in a plane with an acceleration a = (3t2−4)ˆ

i+ 4tˆ

jm/s2. At

t= 0, the particle is at rest at the origin. Find the magnitude and direction of

the particle’s velocity at t= 3 seconds.

Solution

Step 1: To find the particle’s velocity at t= 3 seconds, we first need to find the

particle’s velocity as a function of time by integrating the acceleration function.

v(t) = Za(t)dt =Z(3t2−4)ˆ

i+ 4tˆ

jdt

Integrating each component separately gives:

v(t)=(t3−4t+C1)ˆ

i+ 2t2+C2ˆ

Step 2: We can find the constants C1and C2using the initial conditions at

t= 0. Since the particle is at rest at the origin, we have v(0) = 0.

0 = (03−4(0) + C1)ˆ

i+ 2(0)2+C2ˆ

This gives us C1= 0 and C2= 0, so the velocity function becomes:

v(t) = t3ˆ

i+ 2t2ˆ

Step 3: Now, we can find the velocity of the particle at t= 3 seconds.

v(3) = 33ˆ

i+ 2(3)2ˆ

j= 27ˆ

i+ 18ˆ

Step 4: Finally, we find the magnitude and direction of the particle’s velocity

at t= 3 seconds. The magnitude of the velocity is:

|v(3)|=p(27)2+ (18)2=√729 + 324 = √1053 ≈32.5 m/s

The direction of the velocity can be found using the arctangent function:

Direction = arctan 18

27= arctan 2

3≈33.7◦

Therefore, at t= 3 seconds, the magnitude of the particle’s velocity is ap-

proximately 32.5 m/s and the direction is approximately 33.7 degrees above the

positive x-axis.

Question 34

Question

A particle moves in the xy plane. Its position vector as a function of time

is given by r(t) = (3t2−2t)ˆ

i+ (4t+ 1)ˆ

j, where ˆ

iand ˆ

jare unit vectors in

the xand ydirections, respectively. Determine the magnitude of the particle’s

velocity and acceleration at t= 2 s.

Solution

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

respect to time.

Step 1: v(t) = dr(t)

dt =d

dt (3t2−2t)ˆ

i+ (4t+ 1)ˆ

j

= (6t−2)ˆ

i+ 4ˆ

Step 2: At t= 2 s, we find the velocity vector.

Step 2: v(2) = (6(2) −2)ˆ

i+ 4ˆ

j= 10ˆ

i+ 4ˆ

Step 3: The magnitude of the velocity is given by

Step 3: |v(2)|=p(10)2+ (4)2=√116 = 2√29 m/s

Step 4: Now, we find the acceleration vector by differentiating the velocity

with respect to time.

Step 4: a(t) = dv(t)

dt =d

dt (6t−2)ˆ

i+ 4ˆ

j

= 6ˆ

Step 5: At t= 2 s, we find the acceleration vector.

Step 5: a(2) = 6ˆ

Step 6: The magnitude of the acceleration is given by

Step 6: |a(2)|=|6ˆ

i|= 6 m/s2

Therefore, at t= 2 s, the magnitude of the particle’s velocity is 2√29 m/s

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

Question 35

Question

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

above the horizontal. Find the time it takes for the projectile to reach the highest

point of its trajectory.

Solution

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

The initial velocity of the projectile can be resolved into two components: the

horizontal component (v0x) and the vertical component (v0y).

v0x=v0·cos(θ) = 30 m/s ·cos(60◦)

v0y=v0·sin(θ) = 30 m/s ·sin(60◦)

Step 2: Calculate the time to reach the highest point: At the highest point,

the vertical component of velocity is zero. We can use this fact to find the time

it takes for the projectile to reach this point.

vy=v0y−gt

At the highest point, vy= 0, so:

0 = v0y−gt

Solving for t:

t=v0y

Question 2

Question

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

horizontal with a velocity of 20 m/s. Calculate the maximum height the ball

reaches and the horizontal distance it travels before hitting the ground.

Solution

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

horizontal component of the initial velocity is given by v0x=v0cos(θ) where

v0= 20 m/s and θ= 30◦. The vertical component of the initial velocity is given

by v0y=v0sin(θ).

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

the vertical motion is influenced by gravity, we can use the kinematic equation

vy=v0y−gt where vy= 0 at the maximum height and g= 9.8 m/s2. Solving

for t, we get t=v0y

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

height can be calculated using the kinematic equation y=v0yt−1

2gt2. Substi-

tute v0yand tinto the equation to find the maximum height.

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

twice the time it takes to reach the maximum height, ttotal = 2t.

Step 5: Calculate the horizontal distance traveled. The horizontal distance

can be calculated using the equation x=v0xttotal. Substitute v0xand ttotal into

the equation to find the horizontal distance.

Question 3

Question

A baseball player hits a ball with an initial velocity of 25 m/s at an angle of

30◦above the horizontal. The ball lands on the ground 120 m away. Find 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 initial velocity is given by vix=vicos θ, where

vi= 25 m/s and θ= 30◦.

vix= 25 cos 30◦= 25 ×√3

2≈21.65 m/s

The vertical component of the initial velocity is given by viy=visin θ.

viy= 25 sin 30◦= 25 ×1

2= 12.5 m/s

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

the horizontal component of the velocity. Since the horizontal acceleration is

zero, the horizontal velocity remains constant. Using the equation s=vix·t,

where s= 120 m and vix= 21.65 m/s:

t=s

vix

=120

21.65 ≈5.54 s

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

height can be found using the vertical component of the velocity and the time

of flight. The equation for the maximum height reached is:

h=viy·t−1

2·g·t2

where g≈9.81 m/s2is the acceleration due to gravity. Plugging in the values:

h= 12.5×5.54 −1

2×9.81 ×(5.54)2

h≈35.71 m −153.09 m

h≈ −117.38 m

Since the negative value indicates the ball is below the starting point, the

maximum height reached by the ball during its flight is 117.38 m .

Question 4

Question

A football player kicks the ball with an initial velocity of 20 m/s at an angle of

45◦above the horizontal. Find the maximum height the ball reaches and the

total time the ball is in the air. Assume air resistance is negligible.

Solution

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

The initial velocity (v0) can be resolved into its horizontal component (v0x) and

vertical component (v0y) using trigonometric functions. Given v0= 20 m/s and

the angle θ= 45◦, we have:

v0x=v0cos θ

v0y=v0sin θ

Step 2: Calculate the time to reach maximum height. As the ball reaches

its maximum height, its vertical velocity component becomes zero (vy= 0). We

can use the kinematic equation:

vy=v0y−gt

Substitute vy= 0 and solve for t:

0 = v0sin θ−gt

t=v0sin θ

Step 3: Calculate the maximum height reached. The maximum height (h)

can be found using the kinematic equation:

h=v0yt−1

2gt2

Substitute the values of v0yand tinto the equation and solve for h:

h= (v0sin θ)v0sin θ

g−1

2gv0sin θ

g2

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

ball is in the air is twice the time calculated in Step 2, since the time to reach

maximum height is equal to the time to fall back down. Therefore, the total

time in the air is:

Total time = 2t= 2 v0sin θ

g

Question 5

Question

A baseball is hit with an initial velocity of 30 m/s at an angle of 60◦above the

horizontal. Calculate the maximum height the baseball reaches during its flight.

(Assume no air resistance)

Solution

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

The horizontal component vixis given by:

vix=vicos(θ)

where vi= 30 m/s and θ= 60◦.

vix= 30 cos(60◦) = 30 ×1

2= 15 m/s

The vertical component viyis given by:

viy=visin(θ)

viy= 30 sin(60◦) = 30 ×√3

2= 15√3 m/s

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

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

t=vf−vi

−g

where vi= 15√3 m/s (initial vertical velocity), vf= 0 (final vertical velocity at

maximum height), and g= 9.81 m/s2(acceleration due to gravity).

t=0−15√3

−9.81 ≈1.53 s

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

mum height can be found using the formula:

ymax =viyt−1

2gt2

Substitute viy= 15√3 m/s and t= 1.53 s into the equation:

ymax = 15√3×1.53 −1

2×9.81 ×(1.53)2≈33.9 m

Therefore, the maximum height the baseball reaches is approximately 33.9

meters.

Question 6

Question

A projectile is launched with an initial speed of 50 m/s at an angle of 30 degrees

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

during its flight.

Solution

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

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

follows: The horizontal component: vix=vi·cos(θ) The vertical component:

viy=vi·sin(θ) where vi= 50 m/s is the initial speed, θ= 30◦is the launch

angle.

Substitute the values: vix= 50 ·cos(30◦)viy= 50 ·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 found using the

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

which is −9.81 m/s2.

The vertical displacement yat maximum height is zero. The equation of

motion for vertical motion is: y=viy·t−1

2·g·t2where y= 0, viy= 50·sin(30◦),

g=−9.81 m/s2, and we need to solve for t.

Step 3: Find the maximum height. Once we have the time taken to reach the

maximum height, we can find the vertical distance traveled using the equation:

y=viy·t−1

2·g·t2.

Question 7

Question

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

above the horizontal. The ball lands on the ground 150 meters away. Calculate

the maximum height the ball reaches during its flight.

Solution

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

initial velocity of the ball can be resolved into horizontal and vertical components

as follows: Horizontal component: vix=vicos θ= 30 m/s ×cos 30◦≈25.98 m/s

Vertical component: viy=visin θ= 30 m/s ×sin 30◦= 15 m/s

Step 2: Use the vertical motion equation to find the maximum height. The

equation for vertical motion is: y=viyt−1

2gt2where yis the height, viyis the

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

taken to reach the maximum height.

At the maximum height, the vertical velocity becomes zero. Therefore, we

can find the time taken to reach the maximum height using: 0 = viy−gt Solving

for tgives: t=viy

g=15 m/s

9.8 m/s2≈1.53 s

Now, substitute t= 1.53 s into the vertical motion equation to find the

maximum height ymax:ymax =viyt−1

2gt2= 15 m/s ×1.53 s −1

2×9.8 m/s2×

(1.53 s)2≈11.52 m

Therefore, the maximum height the ball reaches during its flight is approxi-

mately 11.52 meters.

Question 8

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 and the total time of flight. Take g= 9.81 m/s2.

Solution

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

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

follows:

v0x=v0cos(θ)

v0y=v0sin(θ)

Given that v0= 20 m/s and θ= 30◦, we have:

v0x= 20 cos(30◦)≈17.32 m/s

v0y= 20 sin(30◦) = 10 m/s

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

using the vertical component of motion. The projectile reaches its maximum

height when the vertical component of the velocity is zero. Using the equation

for vertical velocity, we have:

vy=v0y−gt

At the maximum height, vy= 0, so:

0 = 10 −9.81t

Solving for t, we get:

t=10

9.81 ≈1.02 seconds

Since at the maximum height, the object spends half of the time of flight, the

total time of flight is given by:

Time of flight = 2t≈2(1.02) ≈2.04 seconds

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

be calculated using the vertical component of motion. We can use the kinematic

equation:

y=v0yt−1

2gt2

Substitute v0y,t, and ginto the equation:

y= 10(1.02) −1

2(9.81)(1.02)2≈5.1 meters

Therefore, the maximum height reached by the projectile is approximately

5.1 meters, and the total time of flight is approximately 2.04 seconds.

Question 9

Question

A ball is thrown with an initial velocity of 15 m/s at an angle of 60 degrees above

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

Solution

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

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

follows:

vix =vicos(θ) = 15 m/s ·cos(60◦)

viy =visin(θ) = 15 m/s ·sin(60◦)

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

vertical velocity at the maximum height is 0, we can use the equation vf=vi+at

for the vertical component to find the time taken to reach the maximum height.

0 = viy −gtmax

tmax =viy

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

using the vertical displacement equation:

∆y=viyt−1

2gt2

∆y=viytmax −1

2g(tmax)2

Substitute the values of viy and tmax:

∆y= 15 m/s ·sin(60◦)15 m/s ·sin(60◦)

9.81 −1

2·9.81 15 m/s ·sin(60◦)

9.81 2

∆y=. . .

(Calculate the final answer to find the maximum height)

Question 10

Question

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

30 degrees above the horizontal. At the same moment, a teammate standing 60

meters away from the player, starts running directly away from the player at a

constant speed of 5 m/s. How far away from the player does the ball land?

Solution

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

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

vix=vi·cos(θ) = 30 m/s ·cos(30◦) = 26 m/s

The vertical component of the initial velocity is given by:

viy=vi·sin(θ) = 30 m/s ·sin(30◦) = 15 m/s

Step 2: Determine the time of flight of the ball. The time of flight can

be found using the vertical motion equation y=viyt+1

2ayt2, where yis the

vertical displacement, viyis the initial vertical velocity, ayis the acceleration in

the y-direction (gravity), and tis the time of flight.

Since the ball lands at the same height it was thrown, y= 0. Therefore the

equation reduces to:

0 = 15t−1

2·9.8 m/s2·t2

9.8t2−15t= 0

t(9.8t−15) = 0

So the time of flight is t= 0 s (at launch) and t=15

9.8≈1.53 s.

Step 3: Determine the horizontal distance the ball travels during this time.

The horizontal distance is given by x=vix·t= 26 m/s ·1.53 s ≈39.78 m.

Step 4: Determine the distance the teammate covers during this time. The

distance the teammate covers is given by dteammate = speed ·t= 5 m/s ·1.53 s =

7.65 m.

Step 5: Determine the total horizontal distance from the player where the

ball lands. The total horizontal distance is the sum of the distance the ball

travels and the distance the teammate covers:

dtotal = 39.78 m + 7.65 m = 47.43 m

Therefore, the ball lands approximately 47.43 meters away from the player.

Question 11

Question

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

above the horizontal. Calculate the maximum height the ball reaches and the

total time the ball is in the air. (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 velocity in the x-direction (v0x) is given by

v0x=v0cos(θ)

v0x= 20 cos(30◦)

v0x≈17.32 m/s

The initial velocity in the y-direction (v0y) is given by

v0y=v0sin(θ)

v0y= 20 sin(30◦)

v0y= 10 m/s

Step 2: Calculate the maximum height (H) the ball reaches. Use the kine-

matic equation for vertical motion:

f=v2

i+ 2a∆y

where vf= 0 (at maximum height) and a=−9.8 m/s2(acceleration due to

gravity).

0 = (10 m/s)2+ 2(−9.8)∆y

0 = 100 −19.6∆y

∆y=100

19.6

∆y≈5.1 m

So, the maximum height the ball reaches is 5.1 meters.

Step 3: Calculate the total time of flight. Use the kinematic equation for

vertical motion:

∆y=v0yt+1

2at2

5.1 = 10t+1

2(−9.8)t2

5.1 = 10t−4.9t2

4.9t2−10t+ 5.1=0

Solving this quadratic equation gives two possible values for t.

t≈1.02 s

t≈1.04 s

Since the total time in the air is the sum of the times it takes to go up and come

down, the total time is approximately 2.06 seconds.

Question 12

Question

A golf ball is struck and launches at an angle of 40◦above the horizontal with an

initial speed of 50 m/s. Calculate the maximum height the ball reaches during

its flight. (Assume air resistance is negligible)

Solution

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

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

components as follows: Vix=Vicos θ= 50 cos 40◦≈38.36 m/s Viy=Visin θ=

50 sin 40◦≈32.11 m/s

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

taken for the golf ball to reach the maximum height can be found using the

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

0 (at maximum height) Using the kinematic equation Vf=Vi+at where a=

−9.8 m/s2: 0 = 32.11 −9.8t t =32.11

9.8≈3.28 s

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

imum height can be calculated using the vertical component of the velocity

at the maximum height and the time taken to reach that height: hmax =

Viyt+1

2(−9.8)t2hmax = 32.11 ×3.28 + 1

2(−9.8)(3.28)2≈52.47 m

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

proximately 52.47 m.

Question 13

Question

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

horizontal with an initial speed of 20 m/s. How far from the player does the

ball land?

Solution

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

The initial velocity (v0) can be resolved into horizontal (v0x) and vertical (v0y)

components: v0x=v0cos(θ) and v0y=v0sin(θ), where v0is 20 m/s and θis 30◦.

Plugging in the values: v0x= 20 cos(30◦)≈17.3 m/s and v0y= 20 sin(30◦)≈

10 m/s.

Step 2: Determine the time of flight. For the vertical motion, we can use

the equation y(t) = v0yt+1

2at2, where ais the acceleration due to gravity (-9.8

m/s

). The ball lands when y(t) = 0. Substitute the known values into the

equation: 0 = 10t−4.9t2. Solving for t, we get two possible solutions: t= 0

(initial time) or t= 2 seconds. Since the time cannot be zero, the time of flight

is 2 seconds.

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

travels (x) can be determined using the equation x=v0x·t. Substitute v0x=

17.3 m/s and t= 2 s into the equation: x= 17.3×2 = 34.6 m.

Therefore, the ball lands approximately 34.6 meters from the player.

Question 14

Question

A ball is launched at an angle of 45◦above the horizontal from the top of a hill

that slopes downward at an angle of 30◦. The magnitude of its velocity is 20

m/s. Calculate the time it takes for the ball to hit the ground.

Solution

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

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

components as follows:

vi,x =vicos α

vi,y =visin α

where viis the magnitude of the initial velocity (20 m/s) and αis the launch

angle (45◦). Substituting the given values into the above equations gives:

vi,x = 20 cos 45◦= 20 ·1

√2= 10√2 m/s

vi,y = 20 sin 45◦= 20 ·1

√2= 10√2 m/s

Step 2: Find the time of flight for the vertical motion. The time of flight

(tflight) for the vertical motion of the ball can be determined using the vertical

motion equation:

y=vi,yt+1

2gt2

Since the ball is launched from the top of the hill, its initial vertical position

(yi) is 0. The final vertical position (yf) is also 0 since the ball hits the ground.

Therefore, the vertical equation simplifies to:

0 = (10√2)t−1

2·9.81 ·t2

10√2t−4.905t2= 0

t(10√2−4.905t)=0

This equation has two solutions: t= 0 and t=10√2

4.905 . Since time cannot be

negative, the time of flight for the vertical motion is t=10√2

4.905 ≈2.03 seconds.

Step 3: Find the time of flight for the horizontal motion. The time of flight

(thill) for the horizontal motion can be determined using the horizontal motion

equation:

x=vi,xthill

The horizontal distance traveled by the ball (x) is determined by the horizontal

component of the ball’s velocity and the time it takes for the ball to hit the

ground. The horizontal distance can be calculated using trigonometry and is

given by:

x=hcot γ

where his the height of the hill and γis the slope angle of the hill (30◦).

Substituting the given values:

x=hcot 30◦=h·√3

Step 4: Combine the horizontal and vertical motions. Since the ball hits the

ground after the same amount of time in both horizontal and vertical motions,

we can equate thill to tflight:

vi,xthill =tflight

10√2thill =10√2

4.905

Solving for thill gives:

thill =1

4.905 ≈0.204 seconds

Therefore, the time it takes for the ball to hit the ground is approximately

0.204 seconds.

Question 15

Question

A particle moves in the xy plane with an acceleration given by a = (4t−2)ˆ

i+6ˆ

where tis in seconds and the particle’s initial velocity is v0= 2ˆ

i+ 3ˆ

jm/s. Find

the magnitude and direction of the particle’s velocity at t= 2 s.

Solution

Step 1: Find the particle’s velocity at time tby integrating the acceleration

function with respect to time.

Za dt =Z(4t−2)ˆ

i+ 6ˆ

j dt

= (2t2−2t)ˆ

i+ 6tˆ

j+

where 

Cis the constant of integration.

Step 2: Use the initial velocity to find the constant of integration 

v0= 2ˆ

i+ 3ˆ

j= (2(0)2−2(0))ˆ

i+ (6(0))ˆ

j+



C= 2ˆ

i+ 3ˆ

Step 3: Substitute the acceleration function and constant of integration into

the velocity function.

v = (2t2−2t)ˆ

i+ 6tˆ

j+ 2ˆ

i+ 3ˆ

Step 4: Determine the velocity at t= 2 s.

v(2) = (2(2)2−2(2))ˆ

i+ 6(2)ˆ

j+ 2ˆ

i+ 3ˆ

= 4ˆ

i+ 12ˆ

j+ 2ˆ

i+ 3ˆ

= 6ˆ

i+ 15ˆ

Step 5: Calculate the magnitude and direction of the particle’s velocity at

t= 2 s.

|v(2)|=p(6)2+ (15)2

|v(2)|=√36 + 225

|v(2)|=√261 m/s

The direction of the velocity can be found by calculating the angle using the

arctangent function.

θ= arctan 15

6

θ≈67.38◦

Therefore, at t= 2 s, the magnitude of the particle’s velocity is √261 m/s at

an angle of 67.38◦above the positive x-axis.

Question 16

Question

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

speed of 40 m/s. Determine the maximum height reached by the projectile.

Solution

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

vertical component is given by vi,y =visin θwhere viis the initial speed (40

m/s) and θis the launch angle (30◦).

Step 2: Calculate the initial vertical component:

vi,y = 40 sin(30◦)≈20 m/s

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

the maximum height. The equation is vf=vi+at, where vfis the final velocity

(0 m/s at the maximum height), viis the initial velocity (20 m/s upwards), a

is the acceleration due to gravity (−9.8 m/s2downward), and tis the time:

0 = 20 −9.8t

Step 4: Solve for the time to reach the maximum height:

t=20

9.8≈2.04 s

Step 5: Use the kinematic equation for vertical motion to find the maximum

height above the launch point. The equation is y=viyt +1

2at2, where yis the

height, viyis the initial vertical velocity, tis the time, and ais the acceleration

due to gravity:

y= 20 ×2.04 + 1

2×(−9.8) ×(2.04)2

Step 6: Calculate the maximum height:

y≈20.4 m

Therefore, the maximum height reached by the projectile is approximately

20.4 meters.

Question 17

Question

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

of 20 m/s. At the highest point of its trajectory, the ball explodes into two

fragments of equal mass. One fragment, initially at rest, falls vertically. The

acceleration of gravity is 9.81 m/s2. What is the speed of the other fragment

just after the explosion?

Solution

Step 1: Calculate the initial velocity components of the ball.



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

vertical components. The horizontal component is vix=vicos(θ), where

vi= 20 m/s and θ= 45◦. Thus, vix= 20 cos(45◦) = 20 ×√2

2= 10√2 m/s.



The vertical component is viy=visin(θ), where vi= 20 m/s and θ= 45◦.

Thus, viy= 20 sin(45◦) = 20 ×√2

2= 10√2 m/s.

Step 2: Calculate the time taken for the ball to reach the highest point of

its trajectory.



The time taken to reach the highest point can be found using the vertical

component of velocity. The formula is vf=vi+at, where vf= 0 at the

highest point and a=−9.81 m/s2. Thus, 0 = 10√2−9.81t. Solving for t,

we get t=10√2

9.81 ≈1.42 s.

Step 3: Calculate the height reached by the ball.



The height reached by the ball can be found using the vertical component

of motion. The formula is y=viyt+1

2at2, where viy= 10√2 m/s, a=

−9.81 m/s2, and t≈1.42 s. Thus, y= (10√2×1.42)+ 1

2×−9.81×(1.42)2≈

10 m.

Step 4: Calculate the velocity of the other fragment just after the explosion.



The velocity of the other fragment after the explosion can be found using

the conservation of momentum. Since the fragments have equal mass, the

vertical component of velocity of the other fragment will be 10√2 m/s

as the original ball. Therefore, the speed of the other fragment just

after the explosion is v=p(vix)2+ (viy)2=q(10√2)2+ (10√2)2=

√200 + 200 = √400 = 20 m/s.

Question 18

Question

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

20 m/s. If air resistance is negligible, how far from the release point does the

ball hit the ground? (Take the acceleration due to gravity as −9.8 m/s2)

Solution

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

components. The initial velocity v0can be resolved into its horizontal (v0x) and

vertical (v0y) components as:

v0x=v0cos(30◦)

v0y=v0sin(30◦)

Step 2: Determine the time tit takes for the ball to hit the ground. Since the

vertical motion is solely under the influence of gravity, we can use the equation

of motion:

y=v0yt+1

2gt2

where yis the vertical displacement and gis the acceleration due to gravity.

Since the ball hits the ground, y= 0. Substituting the values of v0yand g

into the equation, we get:

0 = v0sin(30◦)t−1

2gt2

Solving for t, we get:

t=2v0sin(30◦)

Step 3: Calculate the horizontal distance the ball travels. Since the horizon-

tal motion of the ball is at constant velocity, the distance traveled horizontally

is given by:

x=v0xt

Substitute the values of v0xand tinto the equation:

x=v0cos(30◦)×2v0sin(30◦)

Step 4: Simplify to find the final answer. Calculating the value of x, we get:

x=2v2

0sin(30◦) cos(30◦)

x=v2

0sin(60◦)

Substitute v0= 20 m/s and g= 9.8 m/s2into the equation:

x=202×sin(60◦)

9.8≈34.87 m

Therefore, the ball hits the ground approximately 34.87 meters away from

the release point.

Question 19

Question

A projectile is launched from ground level with an initial velocity of 100 m/s at

an angle of 30◦above the horizontal. Find the horizontal and vertical compo-

nents of its velocity 1 second after launch.