Question:
A ball is dropped from a height of $10 \mathrm{~m}$. If the energy of the ball reduces by $40 \%$ after striking the ground, how much high can the ball bounce back? $\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$. (CBSE 2013)
Solution:
P.E. $=m g h$, Reduced energy $=\frac{60}{100} m g h=\frac{6}{10} m g h$
Let $h^{\prime}=$ height upto which body goes after bounce.
$\therefore \frac{6}{10} m g h=m g h^{\prime}$
$\therefore h^{\prime}=\frac{6}{10} \times 10=6 \mathrm{~m}$