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

Let the ball hit water in t second.

For first ball,         $s=ut+\frac{1}{2}a{t}^2$

                   $122.5=0+\frac{1}{2}\times 9.8\times t^2=4.9t^2$ 

$\Rightarrow$                        $t=\sqrt{\frac{122.5}{4.9}}=\sqrt{25}=5 \mathrm{s}$

For second ball,  $122.5=u(5-2)+\frac{1}{2}\times 9.8\times (5-2{)}^2$

                                     = 3u + 44.1
$\Rightarrow$                           3u = 122.5 ~ 44.1
                                3u = 78.4 $\Rightarrow$ u = $26.1 {\mathrm{ms}}^{-1}$