posted by Cara .
A ball is dropped from the top of a 53.0m high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 25.0m/s The stone and ball collide part way up.
How far above the base of the cliff does this happen in meters?
Write equations for the heights of each ball vs time and set them equal. Solve the resulting equation for time. Then compute the vertical height at that time.
y1 = 53.0 - 4.9 t^2
y2 = 25 t - 4.9 t^2
y2 - y1 = 0 = 25 t - 53
t = 53/25 s = 2.12 s
y1 = 31.0 m