A ball from a bridge 122.5 m above a river.After two seconds, a second ball is thrown straight down after it.

What must be its initial velocity so that both balls hitthe water at the same time?

For 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.