A ball is dropped from a bridge 122.5 metres above a river. After the ball has been falling for two seconds, a second ball is thrown straight down after it. What must its initial velocity be so that both hit the water at the same time

Let f be the time of fall of the first ball to reach the water.Using the formulas=ut+12gt2, we have122⋅5=0+12×9⋅8×t2Solving, we get t2 = 25 or t = 5 secFor the second ball, t = 5 - 2 = 3 sec122⋅5=u×3+12×9⋅8×(3)2Solving it, we get u = 26 .1 m/sec