Please do this
A ball is dropped from a bridge 122.5 m above a river.
After the ball has been falling for two second, a second ball is thrown straight down after it. What must be the initial velocity of the second ball so that both the balls hit the water at the same time?

Dear Student,

​​​​​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.

Regards