Question

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

• A. $10m$
• B. $20m$
• C. $15m$
• D. $30m$

Answer 1

Answer: (B)

Let at the time explosion velocity of one piece of mass is (10 $\hat{i}$ ). If the velocity of other be ${\overrightarrow{v}}_2$ , then from conservation law of momentum (since there is no force in the horizontal direction), the horizontal component of ${\overrightarrow{v}}_2$ , must be $-10$ $\hat{i}$ .
$\therefore$ The relative velocity of two parts in the horizontal direction $=20m{s}^{-1}$
Time taken by the ball to fall through 45 m,
$=t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 45}{10}}=3s$ and time taken by the ball to fall through first 20m, $t^{\prime }=\sqrt{\frac{2h\prime }{g}}=\sqrt{\frac{2\times 20}{10}}=2s$ . Hence time taken by ball pieces to fall from 25 m height to ground $=t-t^{\prime }=3-2=1s$ .
$\therefore$ The horizontal distance between the two pieces at the time of striking on the ground
$=20\times 1=20$ m