A ball is dropped from rest from the top of a cliff is 25.0m high

Question

A ball is dropped from rest from the top of a cliff that is 25.0 m high. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is exactly the same as that with which the first ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just the reverse of each other. Determine how far below the top of the cliff the balls cross paths.

Answer 1

set x=0 at the top of the cliff and x increases as the ball goes down.

 

1. The falling ball hits the ground at

       25 = (1/2)*10* t^2 , note: (s = 0.5gt^2, g = 10)

       t = sqrt(5), this implies the velocity is

       v=gt = 10*sqrt(5)   note: (v0=0)

 

2. the two balls cross when:

    (1/2)*g*t^2 = s = 25 - vt + (1/2)*g*t^2, where v =10*sqrt(5) as indicated in 1 above

    ==> 25 = 10*sqrt(5)*t1

            t1 = (1/2)*sqrt(5)

 

3. the falling ball would have traveled:

          d = (1/2)*10*t1^2 = (1/2) *10*(1/4)*5 = 25/4 meters

 

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