Question

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

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}\\ \\ \\ \sf{Let \:1st\: ball\: hits \:water\: in\: t\: sec}\\ \\ \sf\implies\:S = ut + \dfrac {1}{2} × a×t^{2}\\ \\ \sf\: 122.5 = 0 + \dfrac {1}{2} × 9.8 t^{2} = 4.9t^{2}\\ \\ \sf\:t = \sqrt {\dfrac {122.5}{4.9}} = 5s \\ \\ \sf {For\: second\: ball,} \\ \\ \sf\: 122.5 = u (5 - 2) + \dfrac {1}{2} × 9.8 (5-2)^{2} \\ \\ \sf\: = 3u + 44.1 \\ \\ \sf\implies\: 3u = 78.4 \\ \\ \sf\:u = 26.1 ms^{-1}$

Answer 2

Answer:

The answer is in the pic above.