You shall not spin!

Since there is no external torque, we have $\Delta L = 0$ , thus the system conserves angular momentum. Prior to the collision, $L$ comes entirely from the movement of the ball. After the collision, $L$ comes from the ball-and-rod spinning about the center of mass of the system.

This suggests the center of mass as the simplest point about which we can calculate $L$ .

If $m$ is the mass of the ball, and $M$ the mass of the rod, we have the position of the center of mass as

$r_\text{COM} = \frac{m\times 0 + M \times \ell / 2}{m + M} = \frac{\ell}{2}\frac{M}{m + M}$

The moment of inertia of the ball-and-rod is then

$\begin{aligned} I &= m r_\text{COM}^2 + \frac{M}{\ell}\int\limits_{-r_\text{COM}}^{L-r_\text{COM}} r^2 dr \\ &= \frac{\ell^2 M\left(4m+M\right)}{12\left(m+M\right)} \end{aligned}$

Employing $\Delta L = 0$ , we can say $I\omega = m r_\text{COM} v$ , and thus $\omega = \frac{6mv}{L\left(4m+M\right)}$ .

Now, we want the ball-and-rod to perform one rotation before hitting the ground. It takes the time $\sqrt{2(h-l)/g}$ for the ball-and-rod to hit the ground, thus we have $2\pi = \omega \sqrt{2(h-l)/g}$ .

Solving this relation for $h$ yields $\displaystyle l+\frac{g\ell^2\left(4m+M\right)^2\pi^2}{18m^2v^2}$ .