Given: Distance covered by the ball $s=45 \mathrm{m}$

By using the equation $s=ut+\frac{1}{2}g{t}^2$ where  $u$ represents the initial velocity of ball, $g$ is the acceleration due to gravity, and $t$ is the time taken by the ball.

As the ball is dropped so its initial velocity is zero, thus $s=\frac{1}{2}g{t}^2$

Time taken by the ball to reach the water surface 

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2\times 45}{10}}=3\mathrm{s}$

Speed of boat $=\frac{s}{t}=\frac{12}{3}=4 \mathrm{m} {\mathrm{s}}^{-1}$