projectile motion problems
1. A rock is thrown horizontally at 10.0 m/s from the top of a cliff 122.5 m high.
(a) How long does the rock take to reach the ground?
(b) What is the horizontal displacement of the rock?
2. A ball is thrown from the top of a building with a horizontal velocity of 20 m/s. It hits level ground 80 m from
the face of the building. How high is the building?
3. A player kicks a soccer ball towards the goalkeeper, but at an angle of 37° to the horizontal and with an initial
speed of 14.7 m/s. The goalkeeper stands 26.0 m from the kicker. Where will the ball land relative to the
goalkeeper?
4. A helicopter is rising vertically at a uniform velocity of 14.7 m/s. When it is 196 m from the ground, a ball is
projected from it with a horizontal velocity of 8.5 m/s with respect to the helicopter. Calculate the following.
(a) when the ball will reach the ground
(b) where it will hit the ground
(c) what its velocity will be when it hits the ground
5. A balloon is rising at a vertical velocity of 4.9 m/s. At the same time, it is drifting horizontally with a velocity
of 1.6 m/s. If a bottle is released from the balloon when it is 9.8 m above the ground, determine the following.
(a) the time it takes for the bottle to reach the ground
(b) the horizontal displacement of the bottle from the balloon
6. A cannonball is fired with a velocity of 100 m/s at 25° above the horizontal. Determine how far away it lands
on level ground.
7. A circus clown is fired from a cannon into a net that is situated 2.0 m above the cannon and some distance
from it. The cannon is elevated at 50.0 to the horizontal and the clown’s speed at launch is 15 m/s. See the
diagram below.
(a) Find the horizontal distance from the cannon where the net needs to placed in order for the clown to land
in it.
(b) Calculate the clown’s velocity as he lands in the net.
projectile motion
Answer Section
SHORT ANSWER
1. ANS:
(a) The time required for the projected rock to reach the ground is equal to the time it would take it to fall
straight down. The time required to fall vertically is determined as follows.
Assume that up is positive and down is negative.
(b) The motion in the horizontal direction is uniform.
Thus,
REF: K/U
OBJ: 1.4
LOC: FM1.03
MSC: SP
2. ANS:
Assuming that the horizontal motion is uniform:
KEY: FOP 5.8, p.172
For the vertical motion, assuming up is positive:
REF: K/U
MSC: P
3. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.172
REF: K/U
MSC: P
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.9, p.179
PROBLEM
4. ANS:
(a) Consider first the vertical component of the ball’s motion.
When the ball strikes the ground,
The solutions of the equation are
The time taken to reach the ground is 8.0 s, since the negative solution has no meaning in this problem.
(b)
(c) The vertical component of the velocity is
The horizontal component of the velocity is 8.5 m/s. Therefore the resultant velocity is the vector sum of the
vertical and horizontal components as follows:
The velocity of impact is 64 m/s [82° below the horizontal].
REF: K/U
MSC: SP
5. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.173
(a)
(Negative answer has no meaning in this question.)
(b) Since the balloon and the object are both moving with horizontal velocity of 1.6 m/s, there will be no
horizontal displacement of the object from the balloon. (Relative to the ground there will be a horizontal
displacement.)
REF: K/U, C
MSC: P
6. ANS:
Solution 1:
Solution 2:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.175
REF: K/U
OBJ: 1.4
LOC: FM1.03
MSC: P
7. ANS:
(a)
Time of flight: let “up” be (–) and “down” be (+)
v1 = –15 m/s(sin 50) = –11.5 m/s
a = 9.8 m/s2
d = –2.0 m
t = ?
KEY: FOP 5.9, p.178
–2.0 = (–11.5)t + 4.9(t)2
Solving the quadratic: t = 0.19 s (way up) and 2.16 s (way down)
Horizontal range: d = vt = 15 m/s(cos 50º)(2.16 s) = 21 m
The net must be placed 21 m away from the cannon.
(b) Horizontal component of final velocity: 15 m/s(cos 50) = 9.64 m/s
Vertical component of final velocity: v2 = v1 + at = –11.5 m/s + 9.8 m/s2(2.16 s)
v2 = 9.67 m/s
Using Pythagoras:
=
The shell lands with a velocity of 14 m/s at an angle of 45 below the horizontal.
REF: K/U
OBJ: 1.4
LOC: FM1.03