A ball is dropped from a bridge 122.5 m above a river. After the ball has been falling for 2s, a secondball is thrown straight down after it. What must be the initial velocity of the second ball, so that bothball hit the water at the same time?

Let the ball hit water in t second.For first ball,         s=ut+12at2                   122.5=0+12×9.8×t2=4.9t2 ⇒                        t=122.54.9=25=5 sFor second ball,  122.5=u(5-2)+12×9.8×(5-2)2                                     = 3u + 44.1⇒                           3u = 122.5 ~ 44.1                                3u = 78.4 ⇒ u = 26.1 ms-1