Answer
$e=\sqrt{\frac{h_1}{h_{\circ}}}$
Work Step by Step
We can determine the required coefficient of restitution as follows:
First, we apply the conservation of total energy from position A to position B
$\frac{1}{2}mv_A^2+mgh_{\circ}=\frac{1}{2}mv_{B_1}^2+mgh_B$
$0+mgh_{\circ}=\frac{1}{2}mv_{B1}^2+0$
This simplifies to:
$v_{B1}=\sqrt{2gh_{\circ}}\uparrow$
The conservation of energy when the ball returns from position B to C is:
$\frac{1}{2}mv_B^2+0=\frac{1}{2}mv_C^2+mgh_1$
This simplifies to:
$v_{B2}=\sqrt{2gh_1}\uparrow$
We know that:
$e=\frac{v_{B2}}{v_{B1}}$
We plug in the known values to obtain:
$e=\frac{\sqrt{2gh_1}}{\sqrt{2gh_{\circ}}}$
This simplifies to:
$e=\sqrt{\frac{h_1}{h_{\circ}}}$
