Question

A ball is dropped from a bridge 122.5m above a river.After 2sec ,a second ball is thrown down after it.what must its intial velocity be so that both hit the water at the same time?

Answer 1

or the first ball that is released from the bridge,

Initial velocity, u = 0

Distance travelled, S = 122.5 m

Using,

S = ut + ½ at²

=> 122.5 = 0 + ½ (9.8)t²

=> t = 5 s

For the second ball, let the initial velocity be ‘u’, the time taken to reach below is (5 – 2 =) 3 s and using the same equation,

S = ut + ½ at²

=> 122.5 = 3u + ½ (9.8)(32)

=> u = 26.13 m/s

This is the initial velocity in the downward direction of the second ball.