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?

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 at

u is zero so the equation becomes s = ½ at

64 = ½ x 9.8 x t

t

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 at

u is zero so the equation becomes s = ½ at

29 = ½ x 9.8 x t

t

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 at

u is zero so the equation becomes s = ½ at

21 = ½ x 9.8 x t

t

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?

v

Note that the initial vertical component of the velocity is zero so the equation becomes v

v

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.

speed

speed

speed

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 at

u is zero so the equation becomes s = ½ at

68 = ½ x 9.8 x t

t

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?

v

Note that the initial vertical component of the velocity is zero so the equation becomes v

v

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.

speed

speed

speed

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 at

u is zero so the equation becomes s = ½ at

65 = ½ x 9.8 x t

t

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?

v

Note that the initial vertical component of the velocity is zero so the equation becomes v

v

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.

speed

speed

speed

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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