Question

A ball falls freely from a height of $45 \, m$ . When the ball is at a height of $25 \, m$ , 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. $20 \, m$
• 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 $\overset{ \rightarrow }{v}_{2}$ , then from conservation law of momentum (since there is no force in the horizontal direction), the horizontal component of $\overset{ \rightarrow }{v}_{2}$ , must be $-10$ $\hat{i}$ .
$\therefore $ The relative velocity of two parts in the horizontal direction $=20ms^{- 1}$
Time taken by the ball to fall through 45 m,
$=t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 45}{10}}=3s$ and time taken by the ball to fall through first 20m, $t^{′}=\sqrt{\frac{2 h ′}{g}}=\sqrt{\frac{2 \times 20}{10}}=2s$ . Hence time taken by ball pieces to fall from 25 m height to ground $=t-t^{′}=3-2=1s$ .
$\therefore $ The horizontal distance between the two pieces at the time of striking on the ground
$=20\times 1=20$ m