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?
Physics ?#2 - drwls, Tuesday, September 8, 2009 at 1:44am
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