When on a swing of length $l$ and height $h$ above the ground at rest, swinging to a max angle $\theta_{max}$ from the vertical, at what angle $\theta$ from the vertical should you let go of the swing to maximise the length of time you spend in the air?
Related: Ideal angle to launch from a swing to maximize distance
I derived the travel time in the question you linked:
\begin{align*} t&=\frac{v_y}{g}\left(1+\sqrt{1+\frac{2g(y+h)}{v_y^2}}\right) \\ &=\sqrt{\frac{2\ell}{g}}\left(\xi+\sqrt{\xi^2+1+h/\ell-\cos\theta}\right) \end{align*}
where $\xi=\sin\theta\sqrt{\cos\theta-\cos\theta_m}$
and like before you now take $\frac{\mathrm{d}t}{\mathrm{d}\theta}=0$ and solve for $\theta$
