A ball is dropped from the top of a 50m high cliff. One half second later, a car

Question

A ball is dropped from the top of a 50m high cliff. One half second later, a carefully aim stone is thrown straight up from the bottom of the cliff with a speed of 24 m/s.The stone and the ball collide part way up. How far above the base of the cliff does this happen? A ball is dropped from the top of a 50m high cliff. One half second later, a carefully aim stone is thrown straight up from the bottom of the cliff with a speed of 24 m/s.The stone and the ball collide part way up. How far above the base of the cliff does this happen?

Explanation / Answer

Distance = u*t + 1/2 * a * t^2

Ball
u = 0
D ball = 1/2 * 9.8 * t^2

Stone
Vi = 24 m/s
D stone = 24 * t –1/2 * 9.8 * t^2

D ball + D stone = 50 m
1/2 * 9.8 * t^2 + 24 * t – 1/2 * 9.8 * t^2 = 50

24 t = 50
t = 2.0833 seconds

D ball = 1/2 * 9.8 * 2.0833^2 = 21.27 m
D stone = 24 * 2.0833 – 1/2 * 9.8 * t^2 = 28.73 m
this happen 28.73 m above the base of the cliff  

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