Question

A ball is dropped from a bridge 122.5m high .After the first ball has fall for 2 seconds ,a secind ball is thrown starught down after it,what must ne the initial velocity of the second ball be ,so that both the ball hit hte surface of water at the same timd

Answer 1

height will be same
time reduced to 3 second

Answer 2

Answer:

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.

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