Projectile Motion:- Horizontal Projectiles from a Clifftop

This is an example worksheet with worked solutions to be used with the regenerating worksheets "Clifftop1".

Q.1. Vic fired a gun horizontally from the top of a cliff 64 metres high. The bullet left the barrel of the gun with a speed of 65 m/s.
1 a. How long did it take for the bullet to hit the ground? (nearest 0.1s)
1 b. How far from the base of the cliff did the bullet hit the ground? (nearest m)

Q.2. Rob threw a rotten tomato horizontally from the top of a cliff 29 metres high with an initial speed of 86 m/s.
2a. How long did it take for the tomato to hit the ground? (nearest 0.1s)
2 b. How far from the base of the cliff did the tomato first hit the ground? (nearest m)

Q.3. Ricardo kicked a soccer ball horizontally from the top of a cliff 21 metres high with an initial speed of 78 m/s.
3a. How long did it take for the soccer ball to hit the ground?
3 b. What was the horizontal component of the ball's velocity just before it hit the ground?
3 c. What was the vertical component of the ball's velocity just before it hit the ground?
3 d. What was the speed of the soccer ball just before it hit the ground?
3 e. How far from the base of the cliff did the soccer ball first hit the ground?"

Q.4. Davo threw a stone horizontally from the top of a cliff 68 metres high at a speed of 54 m/s.
4 a. How long did it take for the stone to hit the ground?"
4 b. What was the horizontal component of the stone's velocity just before it hit the ground?
4 c. What was the vertical component of the stone's velocity just before it hit the ground?
4 d. What was the speed of the stone just before it hit the ground?
4 e. How far from the base of the cliff did the stone first hit the ground?

Q.5. Paul fired a slingshot horizontally from the top of a cliff 65 metres high.The pellet left the slingshot with a speed of 51 m/s.
5 a. How long did it take for the pellet to hit the ground?"
5 b. What was the horizontal component of the pellet's velocity just before it hit the ground?"
5 c. What was the vertical component of the pellet's velocity just before it hit the ground?
5 d. What was the speed of the pellet just before it hit the ground?
5 e. How far from the base of the cliff did the pellet first hit the ground?

Worked Solutions

Q.1. Vic fired a gun horizontally from the top of a cliff 64 metres high. The bullet left the barrel of the gun with a speed of 65 m/s.
1 a. How long did it take for the bullet to hit the ground? (nearest 0.1s)

Note that the initial vertical component of the velocity is zero.
s = ut + ½ at2
u is zero so the equation becomes s = ½ at2
64 = ½ x 9.8 x t2
t2 = 64/4.9 = 13.06122
t = √(13.06122) = 3.614 = 3.6 s (nearest 0.1s)

1 b. How far from the base of the cliff did the bullet hit the ground? (nearest m)

Range = horizontal velocity x time of flight
range = 65 x 3.6 = 234m


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Q.2. Rob threw a rotten tomato from the top of a cliff 29 metres high with an initial speed of 86 m/s.
2a. How long did it take for the tomato to hit the ground? (nearest 0.1s)

Note that the initial vertical component of the velocity is zero.
s = ut + ½ at2
u is zero so the equation becomes s = ½ at2
29 = ½ x 9.8 x t2
t2 = 29/4.9 = 5.91837
t = √(5.91837) = 2.433 = 2.4 s (nearest 0.1s)

2 b. How far from the base of the cliff did the tomato first hit the ground? (nearest m)

Range = horizontal velocity x time of flight
range = 86 x = 206.4 = 206m (nearest m)

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Q.3. Ricardo kicked a soccer ball horizontally from the top of a cliff 21 metres high with an initial speed of 78 m/s.
3a. How long did it take for the soccer ball to hit the ground?

Note that the initial vertical component of the velocity is zero.
s = ut + ½ at2
u is zero so the equation becomes s = ½ at2
21 = ½ x 9.8 x t2
t2 = 21/4.9 = 4.28571
t = √(4.28571) = 2.070 = 2.1 s (nearest 0.1s)

3 b. What was the horizontal component of the ball's velocity just before it hit the ground?

There are no horizontal forces acting after the ball has been kicked so the horizontal velocity is unchanged at 78 m/s

3 c. What was the vertical component of the ball's velocity just before it hit the ground?
v2 = u2 + 2 a s
Note that the initial vertical component of the velocity is zero so the equation becomes v2 = 2 a s
v2 = 2 x 9.8 x 21 = 411.6
v = √(411.6) = 20.28793 = 20.3 m/s (nearest 0.1m/s)

3 d. What was the speed of the soccer ball just before it hit the ground?
To add two vectors at a right angle to each other use Pythagoras' theorem.
speed2 = vx2 + vy2
speed2 = 782 + 20.32
speed2 = 6084 + 412.09 = 6496.09 speed = √(6496.09) = 80.598 = 81m/s (nearest m/s)

3 e. How far from the base of the cliff did the soccer ball first hit the ground?"
Range = horizontal velocity x time of flight
range = 78 x 2.1 = 163.8 = 164m (nearest m)

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Q.4. Davo threw a stone horizontally from the top of a cliff 68 metres high at a speed of 54 m/s.
4 a. How long did it take for the stone to hit the ground?"
Note that the initial vertical component of the velocity is zero.
s = ut + ½ at2
u is zero so the equation becomes s = ½ at2
68 = ½ x 9.8 x t2
t2 = 68/4.9 = 13.87755
t = √(13.87755) = 3.725 = 3.7 s (nearest 0.1s)

4 b. What was the horizontal component of the stone's velocity just before it hit the ground?
There are no horizontal forces acting after the stone has been thrown so the horizontal velocity is unchanged at 54 m/s

4 c. What was the vertical component of the stone's velocity just before it hit the ground?
v2 = u2 + 2 a s
Note that the initial vertical component of the velocity is zero so the equation becomes v2 = 2 a s
v2 = 2 x 9.8 x 68 = 1332.8
v = √(1332.8) = 36.50753 = 36.5 m/s (nearest 0.1m/s)

4 d. What was the speed of the stone just before it hit the ground?
To add two vectors at a right angle to each other use Pythagoras' theorem.
speed2 = vx2 + vy2
speed2 = 542 + 36.52
speed2 = 2916 + 1332.25 = 4248.25 speed = √(4248.25) = 65.179 = 65m/s (nearest m/s)

4 e. How far from the base of the cliff did the stone first hit the ground?

Range = horizontal velocity x time of flight
range = 54 x 3.7 = 199.8 = 200m (nearest m)

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Q.5. Paul fired a slingshot horizontally from the top of a cliff 65 metres high.The pellet left the slingshot with a speed of 51 m/s.
5 a. How long did it take for the pellet to hit the ground?"
Note that the initial vertical component of the velocity is zero.
s = ut + ½ at2
u is zero so the equation becomes s = ½ at2
65 = ½ x 9.8 x t2
t2 = 65/4.9 = 13.2653
t = √(13.2653) = 3.642 = 3.6 s (nearest 0.1s)

5 b. What was the horizontal component of the pellet's velocity just before it hit the ground?"
There are no horizontal forces acting after the pellet has been fired so the horizontal velocity is unchanged at 51 m/s

5 c. What was the vertical component of the pellet's velocity just before it hit the ground?
v2 = u2 + 2 a s
Note that the initial vertical component of the velocity is zero so the equation becomes v2 = 2 a s
v2 = 2 x 9.8 x 65 = 1274
v = √(1274) = 35.69314 = 35.7 m/s (nearest 0.1m/s)

5 d. What was the speed of the pellet just before it hit the ground?
To add two vectors at a right angle to each other use Pythagoras' theorem.
speed2 = vx2 + vy2
speed2 = 512 + 35.72
speed2 = 2601 + 1274.49 = 3875.49 speed = √(3875.49) = 62.253 = 62m/s (nearest m/s)

5 e. How far from the base of the cliff did the pellet first hit the ground?

Range = horizontal velocity x time of flight
range = 51 x 3.6 = 183.6 = 184m (nearest m)

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