Let at the time explosion velocity of one piece of mass is (10$\widehat{\mathrm{i}}$). If the velocity of other be ${\overrightarrow{\mathrm{v}}}_2$, then from conservation law of momentum (since there is no force in the horizontal direction), the horizontal component of ${\overrightarrow{\mathrm{v}}}_2$, must be $-10$ $\widehat{\mathrm{i}}$.

$\therefore$ The relative velocity of two parts in the horizontal direction $=\mathrm{20 }\mathrm{m }{\mathrm{s}}^{-1}$

Time taken by the ball to fall through 45 m,

$=t=\sqrt{\frac{2h}{\mathrm{g}}}=\sqrt{\frac{2\times 45}{10}}=3 \mathrm{s}$ and time taken by the ball to fall through first 20m, $t^{\prime }=\sqrt{\frac{2h\prime }{\mathrm{g}}}=\sqrt{\frac{2\times 20}{10}}=\mathrm{2 }\mathrm{s}$. Hence time taken by ball pieces to fall from 25 m height to ground $=t-t^{\prime }=3-2=\mathrm{1 }\mathrm{s}$.

$\therefore$ The horizontal distance between the two pieces at the time of striking on the ground

$=20\times 1=\mathrm{20 }$m