= Solution
At first contact the globally highest trajectory just touches the ceiling at its apex. Its flight time to that point is $t_*=r\omega\cos\theta_*/g$. The horizontal displacement from the wheel axis is
$$
x_{\rm hit}=-r\cos\theta_*+r\omega\sin\theta_*\frac{r\omega\cos\theta_*}{g}=\cos\theta_*\left(-r+\frac{r^2\omega^2}{g}\sin\theta_*\right)=0.
$$
Hence \b[the first point struck is directly above the rotation axis], independently of $h$. The release starts to the left of the axis, but the drop's horizontal travel exactly cancels that offset at the optimum apex. The first-contact <angular velocity> satisfies $r^2\omega_*^2=g(h+\sqrt{h^2-r^2})$.
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