When I was little (and even now if I can get the chance) I liked to play on swings, and my favorite method of dismounting was to let go mid-swing and fly thru the air.
That got me wondering what the best point is to let go of a swing to maximize the landing distance. Naively, one might think the point is where the swing makes a $45^\circ$ angle with the vertical, but that may not be so, because with a swing (unlike a cannon), the launch speed decreases with launch angle. Further complicating matters, the landing distance is short enough to be comparable to the length of the swing, meaning that the mere extra initial distance you get from launching late in a swing (some fraction of the swing's length) could be a significant addition to the distance traveled upon landing.
So the question is: At what point in a swing should a person let go to maximize the landing distance?
Note: I give my own answer to this question below, but it's only one method of tackling the problem. Other answers making use of different assumptions, approximations, techniques, etc. are most certainly welcome! That's part of the reason I left the question statement rather general.
Best Answer
Here is the general solution in the released pendulum model (since you asked me to post it).
Here is the definition of my variables:
Energy conservation gives $\frac{1}{2}v^2+gy=gm$ which gives our total velocity $v$: $$v^2=2g(m-y)$$
Our coordinate system gives us: \begin{align*} x&=\ell\sin\theta \\ y&=\ell-\ell\cos\theta \\ v_x&=v\cos\theta \\ v_y&=v\sin\theta \end{align*}
Considering $y$ kinematics when you land (at vertical coordinate $=-h$) gives you flight time. $$-h=y+v_yt-gt^2/2\implies t=\frac{v_y}{g}\left(1+\sqrt{1+\frac{2g(y+h)}{v_y^2}}\right)$$
The $x$ kinematics give you the distance you travel. \begin{align*} x_\text{land}&=x+v_xt\text{ (and now just plug in and we are done)} \\ &=x+\frac{v_xv_y}{g}\left(1+\sqrt{1+\frac{2g(y+h)}{v_y^2}}\right) \\ &=x+\frac{v^2\sin(2\theta)}{2g}\left(1+\sqrt{1+\frac{2g(y+h)}{v_y^2}}\right) \\ &=x+\sin(2\theta)(m-y)\left(1+\sqrt{1+\frac{2g(y+h)}{v^2\sin^2\theta}}\right) \\ &=x+\sin(2\theta)(m-y)\left(1+\sqrt{1+\frac{y+h}{(m-y)\sin^2\theta}}\right) \\ &=\ell\sin\theta+\sin(2\theta)(m-\ell+\ell\cos\theta)\left(1+\sqrt{1+\frac{\ell-\ell\cos\theta+h}{(m-\ell+\ell\cos\theta)\sin^2\theta}}\right) \end{align*}
The $\theta$ you're interested in is the unique maximum of $x_\text{land}$ satisfying $0\leq\theta\leq\theta_m$. Since $\theta_m$ is the max swing angle, we can relate it to max height $m$ by $m=\ell-\ell\cos\theta_m$. From this you can see the parameter $m$ is just given by the max angle, so you can replace it.
$$x_\text{land}=\ell\sin\theta+\ell\sin(2\theta)(\cos\theta-\cos\theta_m)\left(1+\sqrt{1+\frac{\ell-\ell\cos\theta+h}{\ell(\cos\theta-\cos\theta_m)\sin^2\theta}}\right)$$
You can make this 'unitless' by switching to units where $\ell=1$ (which is how I work on my paper). To go back to units, everywhere you see a variable with units of distance replace it with the same variable divided by $\ell$ to get the unitful equation again.
$$x_\text{land}=\sin\theta+\sin(2\theta)(\cos\theta-\cos\theta_m)\left(1+\sqrt{1+\frac{1-\cos\theta+h}{(\cos\theta-\cos\theta_m)\sin^2\theta}}\right)$$
To use this to find the release angle you pick $h\geq 0$ and $0<\theta_m\leq\pi/2$ then numerically solve $\frac{\mathrm{d}x_\text{land}}{\mathrm{d}\theta}=0$ for $\theta$ subject to the constraint $0\leq\theta\leq\theta_m$.
Here are some plots over all $\theta_m$. I'm plotting the fraction of maximum swing angle you release at, $\theta/\theta_m$.
$h=0$:
$h=\ell/2$:
$h=\ell$:
The system you used is the one when $h=0$ and $\theta_m=\pi/2$.