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Projectile motion is motion that consists of independent horizontal and vertical components.
viewed from above  viewed from the side  

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.
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/s^{2} 
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 = v_{iv}t + (0.5)at^{2} where v_{iv} = 30, t = 4.0 and a = 0 . Solve for d
b. To find the horizontal component of velocity at 4.0 s :
v_{fh} = v_{ih} + at = 30 m/s + 0 = 30 m/s
c. To find the vertical component of the velocity at 4.0 s :
v_{fv} =
v_{iv}
+ at = 0 + (10)(4.0) = 40 m/s
d. The resultant velocity at 4.0 s is
√(30^{2} + 40^{2}) = 50 m/s
ø = tan^{1}(40/30) = 53º below the horizontal
e. The vertical displacement at 4.0 s is:
d = v_{iv}t + (0.5)at^{2} where v_{iv} = 0, t = 4.0 and a = 10 . Solve for d
2. First, find the components of the initial velocity:
v_{iv} =
12sin40º = 7.71 m/s (+ is assigned up)
v_{ih} =
12cos40º = 9.19 m/s
so far, we have:
horizontal  vertical  
distance  ?  ? 
acceleration  0  9.8 m/s^{2} 
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 = ( v_{fv}
 v_{iv})/t
or, t = ( v_{fv}
 v_{iv})/a
t = ( 0  7.71)/(9.8) = 0.7867 s
The
maximum height
is given by
d = v_{iv}t
+ (0.5)at^{2}
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/s^{2} 
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 = v_{ih}t
+ (0.5)at^{2}
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 = ( v_{fh}
 v_{ih})/t
or
v_{fh} =
v_{ih} + at =
9.19 m/s + 0 = 9.19 m/s
In the vertical component,
a = ( v_{fv}
 v_{iv})/t
or,
v_{fv} =
v_{iv} + 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/s^{2}
[down].
________________
PROJECTILE MOTION PROBLEMS:
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 boy kicked a can horizontally from a 6.5 m high rock with a speed of 4.0 m/s. How far from the base of the rock the can land?
5. A pilot flying a constant 215 km/h horizontally in a lowflying 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. A fleck moving horizontally to the right at 2.5 m/s begins to accelerate downward at 0.75 m/s^{2} . Where is the fleck 4.0 s later?
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 ball is kicked from a point 38.9 m away from the goal. The crossbar is 3.05 m high. If 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 30 m/s^{2} near Earth's surface. A bolt separates from the rocket. What is the acceleration of the bolt?
5. Water is leaving a hose at 6.8 m/s. If the target is 2 m away horizontally, What angle should the water have initially?
6. A 5.0 kg brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how tall is the building?
7. A spear is thrown upward from a cliff 48 m above the ground. Given an initial speed of 24 m/s at an angle of 30º to the horizontal,
a. how long is the spear in flight?
b. what is the magnitude and direction of the spear'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 direction of the velocity at the maximum height.
9. A projectile leaves a gun at the same instant that the target is dropped from rest. If the projectile is initially aimed straight at the target, will it hit the target?
10. A basketball is lobbed toward a hoop 3.05 m above the floor. If released 2 m above the floor 10 m from the basket and at a 45 degree angle, how fast must the basketball be thrown so that it goes through the hoop?
11. Dick is tossing chocolates up to
Jane's window from 8.0 m below her window and 9.0 m from the base of the
wall. If the chocolates are traveling horizontally through the open window,
how fast are they going through her window?
12. A projectile has an initial 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 water balloon hits a target 26 m away, at the same height as the release point. The horizontal component of the initial velocity was 5 m/s. What was the vertical component of the initial velocity? What was the launch angle?
15. A soccer ball leaves a cliff 20.2 m above the valley floor, at an angle of 10 degrees above the horizontal. The ball hits the valley floor 3.0 seconds later. What is the initial velocity of the ball? What maximum height above the cliff did the ball reach?
16. A flea stands 2.00 m from a dog's haunches .55m in height. Jumping at an angle of 32 degrees, what initial speed must the flea have to reach her new home?
17. A bullet hit a target 301.5m away. What maximum height above the muzzle did the bullet reach if it was shot at an angle of 25 degree to the ground?
18. A 3.00 kg parcel is dropped out of a window from a height of 176.4 m. Wind exerts an average 12.0 N force on the parcel away from the building. How long is the parcel in the air? Where does it land? What is its impact velocity?
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. the horizontal component of initial
velocity
b. the vertical component of initial
velocity
c. initial speed
20. An arrives 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. the horizontal component of initial
velocity
b. the vertical component of initial
velocity
c. the vertical component of the
impact velocity
d. the horizontal component of the
impact velocity
21. Find the minimum initial speed of a champagne cork that travels a horizontal distance of 11 meters.
22. During practice, a soccer player kicks a
ball, giving it a 32.5 m/s initial speed. It travels the maximum possible
distance before landing down field.
(a) How much time does the ball spend
in the air?
(b) How far did the ball travel?
23. A projectile was launched 64° above the horizontal, attaining a height of 10 m. What is the projectile's initial speed?
24. At what launch angle will the range of a projectile equal its maximum height?
25. A boy kicks a soccer ball directly at a wall 41.8 m away. The ball leaves the ground at 42.7 m/s with an angle of 33.0 degrees to the ground. What height will the ball strike the wall?
26. A golfer on the green imparts a velocity of 6.26 m/s to a ball at an angle of 50° above the ground. The ball travels a distance L through the air, and rolls a distance L at 3.30 m/s before sinking into the hole. What is the ball's average speed throughout the putt?
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 boulder rolls 35 m down a hill, starting from rest and accelerating at 3.06 m/s^{2}. The boulder then rolls off a 45 m high vertical cliff, launching at 19.0° below the horizontal. (a) How far from the cliff's base does the boulder land? (b) How much time does the boulder spend falling?
29. What is the relationship between the maximum height of the projectile, the projectile's range, and the launch angle?
A8.
x

y


distance 
?

?

acceleration 
0

0.75 m/s^{2}

initial velocity 
2.5 m/s

0 m/s

final velocity 


time 
4.0 s

4.0 s

To find the horizontal displacement at 4.0 s:
d = v_{i}t + (0.5)at^{2} where v_{i} = 2.5, t = 4.0 and a = 0 .
d = 10 m
To find the vertical displacement at 4.0 s:
d = v_{i}t + (0.5)at^{2} where v_{i} = 0, t = 4.0 and a = 0.75 m/s^{2}.
d = 6 m
The position of the fleck at 4.0 s is 10 m to the right and , 6 m down from the point the acceleration started.
B28.
To find the magnitude of the velocity of the boulder leaving the cliff, we apply v_{f}^{2} = v_{i}^{2} + 2ad
v_{f}^{2} = 0 ^{2} + 2(3.06)(35.0)
v_{f} = 14.64 m/s
horizontal

vertical


distance 
?

45.0 m

acceleration 
0

9.81 m/s^{2}

initial velocity 
14.64cos19.0°

14.64sin19.0°

final velocity 


time 
t

t

In the vertical component, applying d = v_{i}t + (0.5)at^{2}
45.0 = (14.64sin19.0°) t + (0.5)(9.81)t^{2}
Solving the quadratic for the positive root:
t = 2.58 s
The boulder is in the air for 2.58 s
In the horizontal, applying d = v_{i}t + (0.5)at^{2}
d = (14.64cos19.0°)(2.58) + 0
d = 35.7 m
The boulder lands 35.7 m from the base of the cliff.