Is there any intuition for why bisecting an angle gives the optimal angle to throw a ball? Let's say that I am holding a ball and have zero height, and there is a wall with a height of $h$ that is a distance $d$ away. At what angle should I throw the ball such that getting it over the wall requires minimal speed?
We can actually derive this using calculus. Obviously, the lowest speed that would clear the wall for a given angle would barely clear the wall. If the ball is thrown at an angle $\theta$ from the vertical and just barely clears the wall, then it has fallen by $d \cot\theta - h$ from a straight-line path by the time it clears the wall. The time it takes to fall this much is ${\sqrt\frac{2d \cot\theta - 2h}{g}}.$ In this time, if the ball was thrown at speed $v,$ then it would travel a distance of $d$ horizontally, which is equal to $v \sin\theta{\sqrt\frac{2d \cot\theta - 2h}{g}}.$  Therefore, $d = v \sin\theta{\sqrt\frac{2d \cot\theta - 2h}{g}}.$ Solving for $v$ gives that $v = \frac{d}{\sin\theta{\sqrt\frac{2d \cot\theta - 2h}{g}}}.$ We want to minimize $v,$ so we want to maximize the denominator, which is $\sin\theta{\sqrt\frac{2d \cot\theta - 2h}{g}}$. Therefore, we are attempting to maximize its square, which is  $\sin^2\theta{\frac{2d \cot\theta - 2h}{g}}.$ We can factor out positive constants from the expression since we are trying to maximize it to get that we have to maximize $\sin^2\theta(d \cot\theta - h),$ which is $d\sin^2\theta \cot\theta -  h \sin^2 \theta  = d\sin\theta \cos\theta -  h\sin^2 \theta.$ We can write this in terms of $2\theta$ as $\frac{d}{2}\sin(2\theta) -  \frac{h}{2}(1 - \cos(2\theta)).$ Again, we can remove the factors of $\frac{1}{2}$ and get $d\sin(2\theta) -  h(1 - \cos(2\theta)),$ which we are trying to maximize. Expanding this out gives $d\sin(2\theta) -  h + h\cos(2\theta)).$ Again, since we are trying to maximize this, we can remove the $-h$ term, giving us $d\sin(2\theta) + h\cos(2\theta).$ Differentiating and setting this equal to $0$ gives $d\cos(2\theta) - h\sin(2\theta) = 0.$ Dividing through by $\cos(2\theta)$ gives $d - h\tan(2\theta) = 0.$ Therefore, $h\tan(2\theta) = d,$ so $\tan(2\theta) = \frac{d}{h}.$ If $\alpha$ is the angle that points straight at the top of the wall, then $\theta$ is exactly half of $\alpha.$ Is there any intuitive proof for why the optimal angle to throw a ball is exactly half of the angle from the vertical that points straight at the top of the wall?
 A: starting with
$$x=v\,\cos(\theta)\,t\\
y=v\,\sin(\theta)\,t-g\frac{t^2}{2}$$
solve those equations for $~t~,y~$ you obtain
$$y(x,v,\theta)={\frac {x{v}^{2}\sin \left( 2\,\theta  \right) -g{x}^{2}}{{v}^{2}\cos
 \left( 2\,\theta  \right) +{v}^{2}}}
$$
$$t={\frac {x}{v\cos \left( \theta  \right) }}=t_l$$
from here with
$$y(x=d,v,\theta)=h\quad \Rightarrow\quad~\text{initial velocity } \\
v^2=-\frac 12\,{\frac {g{d}^{2}}{ \left( \cos \left( \theta \right) h-d\sin
 \left( \theta  \right)  \right) \cos \left( \theta  \right) }}
=v_I^2$$
the velocity "that hit the wall"  is
$$v_w=\sqrt{\dot x^2+\dot y^2}\bigg|_{\left(v=v_I~,t=\frac{d}{v_I\,\cos(\theta)}\right)}\\
v_w=\sqrt{-\frac 12\,{\frac {g \left( -4\,\sin \left( \theta \right) dh\cos \left( 
\theta \right) +{d}^{2}+4\, \left( \cos \left( \theta \right) 
 \right) ^{2}{h}^{2} \right) }{\cos \left( \theta \right)  \left( \cos
 \left( \theta \right) h-d\sin \left( \theta \right)  \right) }}
}$$

the velocity  $~v_w~$  is minimum at
$$\frac{d}{d\theta}\,v_w=0\quad\Rightarrow\quad
\theta_m=\arctan\left({\frac {h+\sqrt {{d}^{2}+{h}^{2}}}{d}}\right)\quad\Rightarrow~\text{the initial velocity}\\
v_I^2={\frac {g \left( {d}^{2}+{h}^{2}+h\sqrt {{d}^{2}+{h}^{2}} \right) }{
\sqrt {{d}^{2}+{h}^{2}}}}
$$

