Explanation

(b) If a body falls from height h, then from equation of motion we know that it will hit the ground with a velocity say u=$\sqrt{2gh}$ which is also the velocity of approach here. Now, if after collision it gains a height $h_{1}$ then again by equation of motion $v = \sqrt{2gh}$, which is also the velocity of separation .so, by definition of e,
$e = \sqrt{\frac{2gh_{1}}{2gh}\ }or\ \ h_{1} = e^{2}h$
Given ,h=20 m, e=0.9
$\therefore$ height attained after first bounce
$h_{1} = (0.9)^{2} \times 20$
$= 0.9 \times 0.9 \times 20$
$= 16.2$