1© Manhattan Press (H.K.) Ltd. 2.4 Motion under gravity.

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1© Manhattan Press (H.K.) Ltd. 2.4 Motion under gravity

2 © Manhattan Press (H.K.) Ltd. Free fall 2.4 Motion under gravity (SB p. 93) Free fall – objects fall towards earth with same acceleration (acceleration of free fall or acceleration due to gravity (g)) without resistance g = 9.81 m s -2 Acceleration is constant – equations of motion under uniform acceleration can be used Go to More to Know 3 More to Know 3

3 © Manhattan Press (H.K.) Ltd. Free fall 2.4 Motion under gravity (SB p. 94) (a) upward motion, - positive velocity (upwards) - positive displacement (above A) (b) downward motion, - negative velocity (downwards) - positive displacement (above A) for H, G, F, E - negative displacement (below A) for J Go to Example 4 Example 4

4 © Manhattan Press (H.K.) Ltd. End

5 © Manhattan Press (H.K.) Ltd. Free fall We can experience the free fall motion by playing on the free-fall ride in many amusement parks. On the way to the top, we are pressed against our seats of the ride, which provide a sense of weight. But as we free fall, we fall at the same rate as our seats. Without the force of the seats pressing on us, we are just being acted upon by the force of gravity. Return to Text 2.4 Motion under gravity (SB p. 93)

6 © Manhattan Press (H.K.) Ltd. Q: Q: A ball is thrown vertically upwards from the ground with an initial velocity u. The time taken by the ball to reach a height h is t 1. The ball then takes a further time of t 2 to return to the ground. Find, in terms of t 1, t 2 and g, (a) the initial velocity (u), (b) the height (h), and (c) the maximum height (H) reached by the ball. Solution 2.4 Motion under gravity (SB p. 94)

7 © Manhattan Press (H.K.) Ltd. Solution: Solution: (a) Acceleration (a) = −g Total time taken for the ball to return to the ground (t) = ( t 1 + t 2 ) When the ball returns to the ground, displacement (s) = 0. Using (b) When the ball is at a height h,. 2.4 Motion under gravity (SB p. 94)

8 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d):. (c) At the maximum height, s = H, v = 0 Using v 2 = u 2 + 2as 0 = u 2 − 2gH Return to Text 2.4 Motion under gravity (SB p. 94)