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Physics Help >> Index to Physics Homework Help » Projectile Motion


Projectile motion is motion that consists of independent horizontal and vertical components.

 

Two Projectiles - one dropped, one thrown horizontally
  viewed from above viewed from the side
Projectile paths are shown for one ball dropped and the other ball thrown horizontally. The projectile dropped does not seem to move when viewed from above.  The thrown projectile seems to move in a straight line at constant speed when viewed from above. Both projectiles, when viewed from the side, seem to drop in a straight line down with the same acceleration.
The path of two projectiles, one dropped, and the other thrown horizontally, is depicted above. Viewed from above, the dropped projectile does not seem to move, while the thrown projectile seems to move in a straight line at constant speed. Viewed from the side, both projectiles drop straight down with the same acceleration.

 

 

 

When faced with a projectile motion problem, there are things you must do and things you must remember.

Do
1. determine the horizontal and vertical components of the initial velocity.

Projectile motion problems require that the initial velocity be resolved into horizontal and vertical components.

For an object projected horizontally, the vertical component of initial velocity is zero.

For an object thrown at 20 m/s 53º above the horizontal, the vertical component of the initial velocity is 20sin53º = 16.0 m/s [up], and the horizontal component is 20cos53º = 12.0 m/s.


2. break down the problem into two problems: horizontal and vertical.
3. assign - and + signs appropriately

Remember:
1. acceleration is zero in the horizontal component and g in the vertical.
2. at the top of its rise, the projectile has a vertical velocity = zero.



EXAMPLES of PROJECTILE MOTION PROBLEMS:
1. A golf ball is projected with a horizontal velocity of 30 m/s and takes 4.0 seconds to reach the ground. (Assume g= 10 m/s² and the air resistance is negligible.) Calculate: the height from which the golf ball was projected. The magnitude of the golf balls' vertical velocity component just before hitting the ground. The horizontal velocity component. Resultant velocity just before the object strikes the ground. The horizontal component of the object's displacement.

 

2. Erica kicks a soccer ball 12 m/s at an angle of 40 degrees above the horizontal.
a. What is the ball's maximum height?
b. What is the ball's maximum range?
c. With what velocity does the ball strike the ground?
d. What are the ball's acceleration and velocity at the top of its rise?

 

 

ANSWERS TO EXAMPLES OF PROJECTILE MOTION PROBLEMS:

 

1. assigning "+" to down we have:

  horizontal vertical
distance ? ?
acceleration 0 10 m/s2
initial velocity 30 m/s 0 m/s
final velocity ? ?
time 4.0 s 4.0 s

 

a. To find the horizontal displacement at 4.0 s :

 

d = vivt + (0.5)at2 where viv = 30, t = 4.0 and a = 0 . Solve for d

 

b. To find the horizontal component of velocity at 4.0 s :

 

vfh = vih + at = 30 m/s + 0 = 30 m/s

 

c. To find the vertical component of the velocity at 4.0 s :


vfv = viv + at = 0 + (10)(4.0) = 40 m/s

 

d. The resultant velocity at 4.0 s is

 

Projectile motion has two components which add as vectors to produce the observed velocity and acceleration. √(302 + 402) = 50 m/s

 

ø = tan-1(40/30) = 53º below the horizontal

 

 

 

 

 

 

e. The vertical displacement at 4.0 s is:

 

d = vivt + (0.5)at2 where viv = 0, t = 4.0 and a = 10 . Solve for d

 

 

2. First, find the components of the initial velocity:


viv = 12sin40º = 7.71 m/s (+ is assigned up)


vih = 12cos40º = 9.19 m/s

so far, we have:

  horizontal vertical
distance ? ?
acceleration 0 -9.8 m/s2
initial velocity 9.19 m/s 7.71 m/s
final velocity ? ?
time ? ?



a. We can find the time it takes the ball to rise to the top of its trajectory by assigning zero to the final vertical velocity.

a = ( vfv - viv)/t or, t = ( vfv - viv)/a

t = ( 0 - 7.71)/(-9.8) = 0.7867 s

The maximum height is given by
d = vivt + (0.5)at2


d = 7.71(0.7867) + (0.5)(-9.8)(0.7867)2 = 3.03 m

b. To find the time for the whole trip double the time it takes the ball to go only up. Also the time of travel is the same in both the vertical and horizontal components.

t = 2 x 0.787 s = 1.57 s

Now we have:

  horizontal vertical
distance ? ?
acceleration 0 -9.8 m/s2
initial velocity 9.19 m/s 7.71 m/s
final velocity ? ?
time 1.573 s 1.573 s



The range is the horizontal distance traveled:

d = viht + (0.5)at2


d = 9.19(1.57) + (0.5)(0)(1.57)2 = 14.4 m

c. To find the velocity of the ball as it strikes the ground, we find the final velocity in each component and add them as vectors.

In the horizontal component:

a = ( vfh - vih)/t or

 

vfh = vih + at = 9.19 m/s + 0 = 9.19 m/s

In the vertical component,

a = ( vfv - viv)/t or,

 

vfv = viv + at = 7.71 + (-9.8)(1.573) = -7.71 m/s or 7.71 m/s [down]

Adding the two components together the velocity with which the ball strikes the ground is:

9.19 m/s [horizontal] + 7.71 m/s [down] = 12.0 m/s 40 º below the horizon.

d. At the top of its rise the ball's velocity is 9.19 m/s horizontally and its acceleration is 9.8 m/s2 [down].
________________

 

 

PROJECTILE MOTION PROBLEMS:

 

A Horizontal Projectile Motion

 

1. Erica kicks a soccer ball 12 m/s at horizontally from the edge of the roof of a building which is 30.0 m high.
a. When does it strike the ground?
b. With what velocity does the ball strike the ground?

 

2. A car drives straight off the edge of a cliff that is 54 m high. The police at the scene of the accident note that the point of impact is 130 m from the base of the cliff. How fast was the car traveling when it went over the cliff?

 

3. A ball thrown horizontally at 22.2 m/s from the roof of a building lands 36 m from the base of the building. How tall is the building?

 

4. A tiger leaps horizontally from a 6.5 m high rock with a speed of 4.0 m/s. How far from the base of the rock will she land?

 

5. A pilot flying a constant 215 km/h horizontally in a low-flying helicopter, wants to drop secret documents into his contact"s open car which is traveling 155 km/h in the same direction on a level highway 78.0 m below. At what angle (to the horizontal) should the car be in his sights when the packet is released?

 

6. A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 25 m/s. The landing incline falls off with a slope of 33º.

a. How long is the ski jumper air borne?

b. Where does the ski jumper land on the incline?

 

7. Stones are thrown horizontally with the same velocity. One stone lands twice as far as the other stone. What is the ratio of the height of the taller building to the height of the shorter?

 

8.

.

Projectile motion follows a parabolic path.

From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. The bullet puts a hole in a window of another building and hits the wall that faces the window. (y = 0.60 m, and x = 6.8 m.) Using the data in the drawing, determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window.

 

 

 

 

 

 

 

 

 

 

9. A particle moves at speed of 2.5 m/s in the +x-direction. Upon reaching the origin, the particle receives a continuous constant acceleration of 0.75 m/s2 in the -y-direction. What is the position of the particle 4.0 s later?

 

 

B General Projectile Motion


1. In example 2, if Erica kicked the ball from the edge of the roof of a building which is 30.0 m high.
a. When does it strike the ground?
b. How far from the building does it land?

2 . A daredevil decides to jump a canyon of width 10 m. To do so, he drives a motorcycle up an incline sloped at an angle of 15 degrees. What minimum speed must he have in order to clear the canyon?

 

3. A kicker must kick a ball from a point 38.9 m from the goal, & the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.4 m/s at an angle of 52.2º to the horizontal.

a. By how much does the ball clear or fall short of clearing the crossbar?

b. What is the vertical velocity of the ball at the time it reaches the crossbar?

 

4. A rocket is accelerating vertically upward at 9.8 m/s2 near Earth's surface. it releases a projectile. Immediately after release the acceleration of the projectile is:

 

5. A fire hose held near the ground shoots water at a speed of 6.8 m/s. At what angle should the nozzle point for the water to land 2 m away?

 

6. A .64 kg rock is projected from the edge of the top of a building with an initial velocity of 8.6 m/s at an angle 61º above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 10.1 m from the base of the building. Assume the acceleration of gravity is 9.8 m/s/s. How tall is the building?

 

7. A stone is thrown from the top of the building upward with an initial speed of 24 m/s at an angle of 30º to the horizontal. The height of the building is 48 m.

a. how long is the stone in flight?

b. what is the magnitude and direction of the stone's velocity just before it hits the ground?

 

8. A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37º above the horizontal. Determine the the magnitude and the angle made by the velocity vector with the horizontal at the maximum height.

 

9. In a popular lecture demonstration, a projectile is fired at a falling target. The projectile leaves the gun at the same instant that the target is dropped from rest. Assuming that the gun is initially aimed at the target, show that the projectile will hit the target. (One restriction of this experiment is that the projectile must reach the target before the target strikes the floor.)

10. A 2 m tall basketball player wants to make a goal from 10 m from the basket. If he shoots the ball at a 45 degree angle, at what initial speed must he throw the basketball so that it goes through the hoop, 3.05 m above the floor, without striking the backboard?


11. Romeo is chucking pebbles gently up to Juliet's window, and he wants the pebbles to hit the window with

only a horizontal component of velocity. He is standing at the edge of a rose garden 8.0 m below her window and 9.0 m from the base of the wall. How fast are the pebbles are the pebbles going when they hit her window?

 

12. A projectile is projected from the origin with a velocity of 15.0 m/s at an angle of 30 degrees above the horizontal. What is the location of the projectile 2.0 seconds later?

 

13. If a ball is kicked with an initial velocity of 25 m/s at an angle of 60° above the ground, what is the "hang time"?

 

14. A physics teacher loads a bundle of exam papers to shoot into the town landfill with his catapult. He aims for a point 26 m away, at the same height. The initial velocity in the horizontal is 5 m/s. What is the initial velocity in the vertical direction. What is the launch angle he needs to use?

 

15. A golfer on a cliff 20.2 m above the ocean hits a drive at an angle of 10 degrees, but unfortunately slices it to the right so that it goes into the ocean! If the ball lands in the briny 3.0 seconds after it was hit, what is the initial velocity of the ball? What height (above the cliff) did the ball reach before falling downward into the water?

 

16. Salmon often jumps waterfalls to reach their breeding grounds. Starting 2.00 m from a waterfall of .55m in height, at what minimum speed must a salmon jumping at an angle of 32 degree leave the water to continue upstream?

 

17. A golfer can hit a golf ball a horizontal distance of over 300m on a good drive. What maximum height would a 301.5 m drive reach if it were launched at an angle of 25 degree to the ground? (Hint: At the top of its flight, the ball\'s vertical velocity will be zero.)

 

18. A 3.00 kg ball is dropped from the roof of a building 176.4 m high. While the ball is falling to Earth, a horizontal wind exerts a constant force of 12.0 N on the ball. How long does the ball take to hit the ground, how far from the building does it land, and with what speed does it hit the ground?

 

19. A projectile is shot from the ground at an angle of 60 degrees with respect to the horizontal, and it lands on the ground 5 seconds later. Find:
a. initial hor. vel.
b. initial ver. vel.
c. (total) initial speed

 

20. An object comes down 30m away horizontally and 5m above the point from which it was launched. It reaches this point 3 seconds after it was launched. Find:
a. initial ver. vel.
b. initial hor. vel.
c. Final ver. vel. (before impact)
d. Final hor. vel. (before impact)

 

21. Find the minimum initial speed of a champagne cork that travels a horizontal distance of 11 meters.

 

22. A golfer imparts a speed of 32.5 m/s to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation.
(a) How much time does the ball spend in the air? (s)
(b) What is the longest "hole in one" that the golfer can make, if the ball does not roll when it hits the green? (m)

 

23. The end of a typical launch ramp is directed 64° above the horizontal. With this launch angle, a skier attains a height of 10 m above the end of the ramp. What is the skier's launch speed?

 

24. At what projection angle will the range of a projectile equal its maximum height?

 

25. A firefighter, 41.8 m away from a burning building, directs a stream of water from a ground level fire hose at an angle of 33.0 degrees above the horizontal. If the speed of the stream as it leaves the hose is 42.7 m/s, at what height will the stream of water strike the building?

 

26. A salmon swimming in still water jumps out of the water with a speed of 6.26 m/s at an angle of 50°, sails through the air a distance L before returning to the water, and then swims a distance L under water at a speed of 3.30 m/s before beginning another porpoising maneuver. Determine the average speed of the fish.

 

27. A projectile is fired with an initial velocity of 120 m/s at an angle above the horizontal. If the projectile's initial horizontal speed is 55 m/s, then at what angle was it fired?

 

28. A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 19.0° below the horizontal. The car rolls from rest down the incline with a constant acceleration of 3.06 m/s2 for a distance of 35.0 m to the edge of the cliff, which is 45.0 m above the ocean. (a) Find the car's position relative to the base of the cliff when the car lands in the ocean.( in meters) (b) Find the length of time the car is in the air.(in seconds)

 

29. Prove that Hmax/R=(1/4)tanθ where R = range Hmax = height max

 

30. A dart gun is fired while being held horizontally at a height 1.00 meter above ground level and while it is at rest relative to the ground. The dart from the gun travels a horizontal distance of 5.00 meters. A college student holds the same gun in a horizontal position while sliding down a 45 degree incline at a constant speed of 2.00 m/s. How far will the dart travel if the student fires the gun when it is 1.00 meter above the ground?

 


ANSWERS to PROJECTILE MOTION PROBLEMS:

For solutions to all the problems on this page click here.

 

A9.

A particle moves at speed of 2.5 m/s in the +x-direction. Upon reaching the origin, the particle receives a continuous constant acceleration of 0.75 m/s2 in the -y-direction. What is the position of the particle 4.0 s later?

 

 
x
y
distance
?
?
acceleration
0
-0.75 m/s2
initial velocity
2.5 m/s
0 m/s
final velocity
time
4.0 s
4.0 s

 

To find the x displacement at 4.0 s:

 

d = vit + (0.5)at2 where vi = 2.5, t = 4.0 and a = 0 .

 

d = 10 m

 

To find the y displacement at 4.0 s:

 

d = vit + (0.5)at2 where vi = 0, t = 4.0 and a = -0.75 m/s2.

 

d = -6 m

 

The position of the particle at 4.0 s is (10 m, -6 m)

 

For solutions to all the problems on this page click here.

 

 

B28.

To find the magnitude of the velocity of the car leaving the ramp we apply vf2 = vi2 + 2ad

 

vf2 = 0 2 + 2(3.06)(35.0)

 

vf = 14.64 m/s

 

 
horizontal
vertical
distance
?
45.0 m
acceleration
0
9.81 m/s2
initial velocity
14.64cos19.0°
14.64sin19.0°
final velocity
time
t
t

 

 

In the vertical component, applying d = vit + (0.5)at2

 

45.0 = (14.64sin19.0°) t + (0.5)(9.81)t2

 

Solving the quadratic for the positive root:

 

t = 2.58 s

 

The car is in the air for 2.58 s

 

In the horizontal, applying d = vit + (0.5)at2

d = (14.64cos19.0°)(2.58) + 0

d = 35.7 m

The car lands 35.7 m from the base of the cliff.

 

For solutions to all the problems on this page click here.