A ball falls on the ground from a height of $2.0\text{m}$ and rebounds up to a height of $1.5\text{m}$. The coefficient of restitution for the collision of the ball with the ground is

Take $g=10{\text{m/s}}^2$.

Velocity of ball just before collision with ground
$v_1=\sqrt{2g{h}_1}$ where $h_1=2\text{m}$
$\Rightarrow v_1=\sqrt{2\times 10\times 2}=\sqrt{40}\text{m/s}$
Let the velocity of ball just after collision with ground be $v_2$ upwards, and it reaches a maximum height of $h_2=1.5\text{m}$ (where $v=0$)

Considering downward direction as $+ve$,
$v^2=u^2+2as$ where $s=-h_2,u=-v_2,a=+g$
$0^2=(-v_2{)}^2+2\times (+g)\times (-h_2)$
$\Rightarrow v_2=\sqrt{2g{h}_2}=\sqrt{2\times 10\times 1.5}$
$\therefore v_2=\sqrt{30}\text{m/s}$

For the collision of ball with ground, we can assume ground to be stationary,
$\text{Speed of approach}=v_1$
$\text{Speed of separation}=v_2$

Coefficient of restitution, $e=\frac{\text{Speed of separation}}{\text{Speed of approach}}$
$\Rightarrow e=\frac{v_2}{v_1}=\frac{\sqrt{30}}{\sqrt{40}}$
$\therefore e=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$