Explanation

(b) Let at the time explosion velocity of one piece of mass $m/2$ is (10$\widehat{i}$). If velocity of other be ${\overrightarrow{v}}_{2}$, then from conservation law of momentum (since there is no force in horizontal direction), horizontal component of ${\overrightarrow{v}}_{2}$, must be $- 10$ $\widehat{i}$.
$\therefore$ Relative velocity of two parts in horizontal direction $= 20ms^{- 1}$
Time taken by ball to fall through 45m,
$= 20 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 45}{10}} = 3s$ and time taken by ball to fall through first 20m, $t^{'} = \sqrt{\frac{2h'}{g}} = \sqrt{\frac{2 \times 20}{10}} = 2s$. Hence time taken by ball pieces to fall from 25 m height to ground $= t - t^{'} = 3 - 2 = 1s$.
$\therefore$ Horizontal distance between the two pieces at the time of striking on ground
$= 20 \times 1 = 20$m