= Solution
The coordinate $\psi$ is cyclic, so $p_\psi=C(\dot\psi+\dot\phi\cos\theta)$ is conserved. Initially the spin is zero, hence $\boxed{\dot\psi+\dot\phi\cos\theta=0}$ throughout. The coordinate $\phi$ is also cyclic. After imposing this zero spin, its conserved momentum becomes $(A+mh^2)\dot\phi\sin^2\theta$. Initially $\theta=\pi/2$ and $\dot\phi=\omega$, so $\boxed{\dot\phi\sin^2\theta=\omega}$.
The equation for $z$ is
$$
\frac d{dt}\left[m(\dot z-h\sin\theta\dot\theta)\right]=-mg.
$$
The quantity in parentheses is the vertical velocity of the <centre of mass>. Initially that velocity and its height are both zero: the initial <rotation> is horizontal and the <vertex> starts at rest at the reference height. Twice integrating gives
$$
\boxed{z+h\cos\theta=-\frac12gt^2.}
$$
Independently, the frictionless vertical wire exerts only horizontal force. <Gravity> is therefore the only vertical external force on the body, and the <centre of mass> undergoes vertical free fall. This gives the same equation without the angular equations.
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