Two Dimensional Kinematics
Projectile Motion Problems
1.
Name: _______________________________
One of the fastest recorded pitches in major-league baseball, thrown by
Billy Wagner in 2003, was clocked at 101.0 mi/h. If a pitch were thrown
horizontally with this velocity, how far would the ball fall vertically by the
time it reached home plate, 60.5 ft away?
(2.68 ft)
v0x  101 m ih  45.1 m s and x  60.5 ft 18.4 m
The time to reach home plate is t
x
18.4 m

 0.408 s
v0x 45.1 m s
In this time interval, the vertical displacement is
y  v0yt


1 2
1
2
ayt  0  9.80 m s2  0.408 s   0.817 m
2
2
 3.281 ft
Thus, the ball drops vertically 0.817 m 
  2.68 ft
 1m 
2.
A peregrine falcon is the fastest bird, flying at a speed of 200 mi/h. Nature
has adapted the bird to reach such a speed by placing baffles in its nose to
prevent air from rushing in and slowing it down. Also, the bird’s eyes
adjust their focus faster than the eyes of any other creature, so the falcon
can focus quickly on its prey. Assume that a peregrine falcon is moving
horizontally at its top speed at a height of 100 m above the ground when
it brings its wings into its sides and begins to drop in free fall. How far
will the bird fall vertically while traveling horizontally a distance of 100 m?
The constant horizontal speed of the falcon is
vx  200
m i 0.447 m s

  89.4 m s
h  1 m ih 
The time required to travel 100 m horizontally is t
x
100 m

 1.12 s. The vertical displacement
vx 89.4 m s
during this time is
y  v0yt
1 2
1
2
ayt  0   9.80 m s2   1.12 s  6.13 m
2
2
or the falcon has a vertical fall of 6.13 m
(6.13 m)
3.
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of
18.0 m/s. The cliff is 50.0 m above a flat, horizontal beach. How long after being released does the stone
strike the beach below the cliff? With what speed and angle of impact does the stone land?
(36
We find the time of fall from y  v0yt
t
2 y
a

2  50.0 m

 9.80 m s
2
1 2
ayt with v0y  0 :
2
3.19 s
At impact, vx  v0x  18.0 m s, and the vertical component is
vy  v0y  ayt 0    9.80 m s2   3.19 s   31.3 m s
Thus, v  vx2  vy2 
18.0 m s   31.3 m s
2
2
 36.1 m s
 vy 
3
1  31.
  tan 
  60.1
 18.0 
 vx 
and
  tan1 
or
v  36.1 m s at60.1 below the horizontal
4.
The best leaper in the animal kingdom is the puma, which can jump to a height of
12 ft when leaving the ground at an angle of 45°. With what speed, in SI units,
must the animal leave the ground to reach that height?
(
At the maximum height vy  0, and the time to reach this height is found from
vy  v0y  ayt as t
vy  v0y
ay

0  v0y
g

v0y
g
The vertical displacement that has occurred during this time is
 
 ym ax  vy
 vy  v0y   0  v0y   v0y  v0y
t 

 t 

av
2   2   g  2g

2
 1m 
Thus, if  ym ax  12 ft
  3.7 m , then
 3.281 ft
v0y  2g ym ax  2 9.80 m s2   3.7 m   8.5 m s
and if the angle of projection is   45 , the launch speed is
v0 
v0y
sin

8.5 m s
 12 m s
sin 45
5.
Tom the cat is chasing Jerry the mouse across the surface of a table 1.5 m above the floor. Jerry steps out
of the way at the last second, and Tom slides off the edge of the table at a speed of 5.0 m/s. Where will
Tom strike the floor, and what velocity components will he have just before he hits?
The time of flight for Tom is found from y  v0yt
t
2 y
a

2  1.5 m

 9.80 m s2
1 2
ayt with v0y  0 :
2
 0.55 s
The horizontal displacement during this time is
x  v0xt  5.0 m s  0.55 s  2.8 m
Thus, he lands 2.8 m from the base ofthe table
The horizontal component of velocity does not change during the flight, so vx  v0x  5.0 m s . The
vertical component of velocity is found as
vy  v0y  ayt 0    9.80 m s2   0.55 s   5.4 m s
/s)
6.
An artillery shell is fired with an initial velocity of 300 m/s at 55.0° above the horizontal. To clear an
avalanche, it explodes on a mountainside 42.0 s after firing. What are the x- and y-coordinates of the
shell where it explodes, relative to its firing point?
The horizontal displacement at t 42.0 s is
x  v0xt  v0 cos  t  300 m s  cos55.0  42.0 s  7.23 103 m
The vertical displacement at t 42.0 s is
y  v0yt
1 2
1
ayt   v0 sin  t gt2
2
2
  300 m s  sin 55.0  42.0 s 
1
2
9.80 m s2   42.0 s  1.68  103 m

2
7.
A brick is thrown upward from the top of a building at an angle of 25° to the horizontal and with an
initial speed of 15 m/s. If the brick is in flight for 3.0 s, how tall is the building?
We choose our origin at the initial position of the projectile. After 3.00 s, it is at ground level, so the
vertical displacement is y   H
To find H, we use y  v0yt
1 2
ayt , which becomes
2
 H   15 m s  sin 25  3.0 s 
1
2
9.80 m s2   3.0 s , or H  25 m

2
8.
A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the
horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The
car rolls from rest down the incline with a constant acceleration of 4.00 m/s2 for a distance of 50.0 m to
the edge of the cliff, which is 30.0 m above the ocean. Find (a) the car’s position relative to the base of
the cliff when the car lands in the ocean and (b) the length of time the car is in the air.
(32.5 m, 1.78 s)
The speed of the car when it reaches the edge of the cliff is
v  v02  2a x  0  2 4.00 m s2   50.0 m   20.0 m s
Now, consider the projectile phase of the car’s motion. The vertical velocity of the car as it reaches the
water is
vy   v02y  2ay  y     20.0 m s sin24.0  2  9.80 m s2    30.0 m
2
or
vy   25.6 m s
(b) The time of flight is
t
vy  v0y
ay

 25.6 m s    20.0 m s sin24.0
 1.78 s
 9.80 m s2
(a) The horizontal displacement of the car during this time is
x  v0xt  20.0 m s cos24.0 1.78 s  32.5 m

9.
A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The
projectile lands on a hillside 4.00 s later. Neglect air friction. (a) What is the projectile’s velocity at the
highest point of its trajectory? (b) What is the straight-line distance from where the projectile was
launched to where it hits its target?
(a)
At the highest point of the trajectory, the projectile is moving horizontally with velocity
components of vy  0 and
vx =v0x =v0 cos   60.0 m /s cos30.0  52.0 m s
(b) The horizontal displacement is x  v0xt  52.0 m s  4.00 s  208 m and, from
y   v0 sin  t
1 2
ayt , the vertical displacement is
2
y   60.0 m s  sin 30.0  4.00 s 
1
2
9.80 m s2   4.00 s  41.6 m

2
The straight line distance is
d
2
2
2
 x   y   208 m    41.6 m   212 m
2
10.
A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 100 m/s. The
rocket moves for 3.00 s along its initial line of motion with an acceleration of 30.0 m/s2. At this time, its
engines fail and the rocket proceeds to move as a projectile. Find (a) the maximum altitude reached by
the rocket, (b) its total time of flight, and (c) its horizontal range.
(1.52 x 103 m, 36.1 s, 4.05 x 103 m)
The distance, s, moved in the first 3.00 seconds is given by
1
1
2
s v0t at2   100 m s  3.00 s   30.0 m s2   3.00 s  435 m
2
2
At the end of powered flight, the coordinates of the rocket are:
x1  scos53.0  262 m ,and y1  ssin 53.0  347 m
The speed of the rocket at the end of powered flight is
v1  v0  at 100 m s  30.0 m s2   3.00 s  190 m s
so the initial velocity components for the free-fall phase of the flight are v0x  v1 cos53.0  114 m s and
v0y  v1 sin53.0  152 m s
(a) When the rocket is at maximum altitude, vy  0. The rise time during the free-fall phase can be
found from vy  v0y  ayt as
trise 
0  v0y
ay

0  152 m
 15.5 s
 9.80 m s2
The vertical displacement occurring during this time is
 vy  v0y 
 0  152 m s
3
y  
 trise  
 15.5 s  1.17  10 m
2 
2



The maximum altitude reached is then
H  y1  y  347 m  1.17  103 m  1.52 103 m
(b) After reaching the top of the arc, the rocket falls 1.52 103 m to the ground, starting with zero
1
vertical velocity v0y  0 . The time for this fall is found from y  v0y t ayt2 as
2

tfall 
2 y
ay


2 1.52 103 m
2
-9.80 m s
  17.6 s
The total time of flight is
t tpowered  trise  tfall   3.00  15.5  17.6 s  36.1 s
(c)
The free-fall phase of the flight lasts for t2  trise  tfall  15.5  17.6 s 33.1 s
The horizontal displacement occurring during this time is
x  v0x t2   114 m s  33.1 s  3.78 103 m
and the full horizontal range is
R  x1  x  262 m  3.78 103 m  4.05 103 m
11.
A daredevil decides to jump a canyon. Its walls are equally high and 10 m apart. He takes off by
driving a motorcycle up a short ramp sloped at an angle of 15°. What minimum speed must he have in
order to clear the canyon?
The time to reach the opposite side is t
x
10 m

v0x v0 cos15
When the motorcycle returns to the original level, the vertical displacement is y  0 . Using this in the
relation y  v0y t
1 2
ayt gives a second relation between the takeoff speed and the time of flight as:
2
0   v0 sin15  t


g
1
  g t2 or v0  
t
2
 2sin15 
Substituting the time found earlier into this result yields the required takeoff speed as
v0 
 9.80 m
s2   10 m

 14 m s
2 sin15  cos15 
12.
A home run is hit in such a way that the baseball just clears a wall 21 m high, located 130 m from home
plate. The ball is hit at an angle of 35° to the horizontal, and air resistance is negligible. Find (a) the
initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components
and the speed of the ball when it reaches the wall. (Assume that the ball is hit at a height of 1.0 m above
the ground.)
(a) The time to reach the fence is t
x
130 m
159 m


v0x v0 cos35
v0
At this time, the ball must be 20 m above its launch position.
y  v0y t
1 2
ayt gives
2
 159 m 
2  159 m 
20 m   v0 sin35  
   4.90 m s  

 v0 
 v0 
2
From which, v0  42 m s
(b) From above, t
(c)
159 m
159 m

 3.8 s
v0
42 m s
vx  v0x   42 m s cos35  34 m s
vy  v0y  ayt  42 m s sin35   9.80 m s2   3.8 s   13 m s
v  vx2  vy2 
37 m/s)
 34.1 m s   13.4 m s
2
2
 37 m s
13.
In a very popular lecture demonstration, a projectile is fired at a
falling target as in the figure. 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.)
Choose an origin where the projectile leaves the gun and let the ycoordinates of the projectile and the target at time t be labeled
yp and yT , respectively.
Then,
 yp  yp  0   v0 sin0  t
 yT  yT  h  0 
g 2
t , and
2
g 2
g
t or yT  h  t2
2
2
The time when the projectile will have the same x-coordinate as the
x0
x
target is t

v0x v0 cos 0
For a collision to occur, it is necessary that yp  yT at this time, or

g 2
x0  g 2
h
  t  h  t which reduces to tan0 
2
x0
 v0 cos0  2
 v0 sin0  
This requirement is satisfied provided that the gun is aimed at the initial location of the target. Thus, a
collision is guaranteed if the shooter aims the gun in this manner.
14.
One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while
your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before
or at the same time as the first one. Assume that both snowballs are thrown with a speed of 25.0 m/s.
The first is thrown at an angle of 70.0° with respect to the horizontal. (a) At what angle should the
second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should
the second snowball be thrown after the first in order for both to arrive at the same time?
1 2
ayt to the vertical motion of the first snowball gives
2
1
0   25.0 m s sin70.0 t1    9.80 m s2  t12 which has the non-zero solution of
2
2 25.0 m s sin70.0
t1 
 4.79 s as the time of flight for this snowball.
9.80 m s2
(a) Applying y  v0yt
The horizontal displacement this snowball achieves is
x  v0xt1   25.0 m s cos70.0  4.79 s  41.0 m
Now consider the second snowball, also given an initial speed of v0  25.0 m s, thrown at angle ,
1
and is in the air for time t2 . Applying y  v0yt ayt2 to its vertical motion yields
2
1
0   25.0 m s sin  t2   9.80 m s2  t22
2
which has a non-zero solution of
2 25.0 m s sin
t2 
  5.10 s sin
9.80 m s2
We require the horizontal range of this snowball be the same as that of the first ball, namely
x  v0xt2   25.0 m s cos   5.10 s sin   41.0 m . This yields the equation
41.0 m
 0.321
 25.0 m s  5.10 s
From the trigonometric identity sin 2 =2sin cos , this result becomes
sin cos 
sin2 =2 0.321  0.642, so 2  40.0
and the required angle of projection for the second snowball is
  20.0 above the horizontal
(b) From above, the time of flight for the first snowball is t1  4.79 s and that for the second snowball
is t2   5.10 s sin   5.10 s sin20.0  1.74 s.
Thus, if they are to arrive simultaneously, the time delay between the first and second snowballs
should be
t t1  t2  4.79 s 1.74 s= 3.05 s