Question

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?

Answer 1

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.