Le Fevre High School
SACE Stage 2 Physics
Projectile Motion 1 Solution
1. How long does it take a car to travel 30 metres if it accelerates from rest at 2.0ms-1 ?
s  30m
v1  0ms 1
a  2ms 2
t ?
s  v1t  1 2 at 2
30  0t  1 2 2t 2
t 2  30
t  30  5.5s
2. What is the stopping distance of a car travelling at 28ms-1 (100km/h=1.0 ×102 km/h) if the
reaction time of the driver is 0.50 seconds and then the driver applies the brakes with the car
decelerating at a rate of 6.0ms-2 ?
(a) Reaction Distance
s?
v0  28ms 1
a  0ms -2
t  0.5s
s  v0 t
s  28  0.5
s  14 m
(b) Braking Distance
s?
v0  28ms 1
a  -6ms -2
v  0ms -1
v 2  v02  2as
0 2  28 2  2  (6)  s
s  65m
(c) Total Braking Distance
s  (reaction distance)  (braking distance)
s  14  65
s  79m
3. Suppose a ball is dropped from a 70m high tower. How far will it have fallen after 3.0s?
v1  0ms 1 a  g  9.8ms -2
s  ? t  3.0s
s  v1t  1 2 at 2
s  0  3.0  1 2  9.8  (3.0) 2
s  44m
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4. A person throws a ball upwards into the air with an initial velocity of 15ms-1. Calculate how high
it goes and how long the ball is in the air before it comes back to the throwers hand.
(a) Time of Flight
v1  15ms 1
g  9.8ms 2
v 2  0ms 1 (at max height)
t ?
v 2  v1  at
0  15  (9.8)t
t  1.53s
 time of flight  2  1.53s  3.06s
(b) Height
s  v1t  1 2 at 2
s  15  1.53  1 2  (9.8)  (1.53) 2
s  22.95  11.47
s  11.48m
 The height of the throw is 11.48m
5. A bird can fly at 30km/h. How long does it take to fly 235km?
v  30kmhr 1  8.33ms 1
s  235km  2.35  10 5 m
s
t
s
t
v
2.35  10 5
t
8.33
t  2.82  10 4 s
v
6. A rock thrown horizontally at a large bell 50m away is heard to hit the bell 4.5s later. What was
the speed of the rock? (neglect the effect of gravity and time taken for the sound to travel to the
observer).
t  4.5s
s  50m
s
t
50
v
4.5
v  11.1ms 1
v
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7. At high speeds, a particular car is capable of an acceleration of about 0.50ms-2. At this rate how
long does it take to accelerate from 90km/h to 100km/h?
a  0.5ms 2
v1  90kmhr 1  25ms 1
v 2  100kmhr 1  27.8ms 1
v 2  v1  at
v 2  v1
a
27.8  25
t
0.5
t  5.56s
8. An advertisement claims that a sports car is able to stop, from a speed of 100km/h, within 45m.
What is its acceleration in ms-2? How many “g’s” is this?
v1  100kmhr 1  27.8ms 1
v 2  0ms 1
s  45m
t
v12  v 22  2as
v12  v 22
a
2s
(27.8) 2  (0) 2
a
2  45
a  8.59ms  2
This is equivalent to 8.59
9.8  0.88 g' s
9. A stone is dropped from the top of a cliff. It is seen to hit the ground below after 5.5s. How high
is the cliff?
v1  0ms1
t  5.5s
s  v1t  1 2 at 2
s  0  5.5  1 2  (9.8)  5.5
s  148m
10. A trout jumps a waterfall 2.5m high. With what minimum speed did it leave the water below to
reach the top?
s  2.5m
v1  ?
v 2  0ms 1
a  9.8ms 1
v 22  v12  2as
0 2  v12  2  (9.8)  2.5
v1  7ms 1
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11. When 500m from a station and travelling at 60kmh-1, a train driver notices a stationary train on
the same track at the platform. The brakes are applied and the train decelerates uniformly at 0.3ms-2.
What happens?
a  0.3ms 1
s  500m
v1  60kmhr 1  16.67 ms 1
v 2  0ms 1
v 22  v12  2as
0  (16.67) 2  2  (0.3)  s
s  463m
The train stops comfortably before the stationary train at the platform.
12. Two cars, each travelling at the speed limit of 60kmh-1, collide head on. This is equivalent to a
car being dropped from what height? (assume a=10ms-2)
v  60kmhr 1 Reletive Speed  v1  120kmhr 1  33.34ms 1 v 2  0ms 1
v 22  v12  2as
v 22  v12
2a
2
0  (33.34) 2
s
2  10
s  56m
13. What will the resultant path be of a ship travelling due North at 10ms-1 if there is a tide of 3ms-1
towards the east?
s
AC  resultant vector  v1  v 2
AC  10.44ms 1 at 17 o True
14. A stone dropped from the top of a cliff is heard to hit the water below 5 seconds later. Calculate:
v1  0ms 1
a  9.8ms 1
t  5s
(a) the velocity of the stone just before it hits the water.
v 2  v1  at
v 2  0  9. 8  5
v 2  49ms 1
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(b) its average velocity during this time.
v  v2
v avg  1
t
0  49
v avg 
2
v avg  24.5ms 1
(c) the height of the cliff (g=9.8ms -2 )
s
v avg 
s  v1t  1 2 at 2
t
s  v avg t
or
s  0  5  1 2  9.8  5 2
s  122.5m
s  24.5  5
s  122.5m
15. The photograph below represents an object launched at the Earth’s surface, with negligible air
resistance. The ball has an actual size of 10 cm and the time between each image was 0.02 s.
(a) Use the diagram to verify that the horizontal component of the velocity remains constant.
From the diagram it can be seen that the horizontal displacements
remain approximately equal. Hence there is no change in the
horizontal motion and thus in the component of motion in this
direction.
(b) Use the diagram to estimate the angle at which the projectile is launched.
Considering the two initial images and drawing a tangent to these
images the initial angle appears to be approximately 63o.
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(c) Estimate the initial vertical component of the launch velocity.
Distance between first two images = 1.0 cm. Photographic size of
ball = 0.5 cm, actual size = 10 cm  Scale factor 1 cm: 0.2 m.
Since the time between images = 0.2 s then
v
s 0.2

 10 m s 1 @
t 0.02
vertically upwards component.
(d) State the velocity at which the object lands.
Initial horizontal component of the velocity :
vH 
0. 1
 5 m s 1 directed
0.02
5 m s-1

to the right
10 m
s-1
Initial velocity using Pythagoras
v  10 2  5 5
= 11.2 m s-1 tan  
10
2
5
  = 63o to the horizontal .
It will land with the same speed as the initial speed, but with a
change of 180o in direction.
i.e. 11.2 m s-1 @ 117o clockwise from the horizontal.