= Solution
With zero <gravitational potential energy> at the wheel axis, the release <energy> and the <energy> at the apex of the <projectile motion> are
$$
E_{\rm release}=\frac12mr^2\omega^2+mgr\sin\theta,\qquad E_{\rm apex}=\frac12mr^2\omega^2\sin^2\theta+mgy_{\max}.
$$
Only the vertical <velocity> vanishes at the apex: the horizontal <kinetic energy> remains. Equating the two <energies> gives
$$
\boxed{y_{\max}(\theta)=r\sin\theta+\frac{r^2\omega^2}{2g}\cos^2\theta.}
$$
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