 Solution

The azimuthal coordinate is cyclic, so its canonical momentum is conserved. Time independence of the Lagrangian also conserves the energy. The two constants areFor a horizontal launch from depth $D$, take $0<D<\ell$, so $\cos {\theta }_0=D/\ell$, ${\dot{\theta }}_0=0$, and $v=\ell \sin {\theta }_0{\dot{\varphi }}_0$. Thus $h=mv\sqrt{{\ell }^2-D^2}$ and $E=\frac{1}{2}m{v}^2-mgD$. The meridional motion has effective potential $U(\theta )=h^2/(2m{\ell }^2{\sin }^2\theta )-mg\ell \cos \theta$.