Post a New Question

Physics ?#2

posted by .

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 -

    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

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question