A ball falls freely from a height of $45 \mathrm{m}$. When the ball is at a height of $25 \mathrm{m}$, it explodes into two equal pieces. One of them acquires an additional horizontal component of velocity equal to $10$ $\mathrm{m} {\mathrm{s}}^{-1}$, while its vertical component remains the same. The distance between the two pieces when both strike the ground is

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