Maximum horizontal distance of a freefalling ball with one allowed bounce anywhere along its initial path? I was washing a spoon in the sink and this question popped into my mind:
If a ball is dropped from height H and is allowed a single deflection of any angle $0 < \theta < 180$ at any height $ 0 \leq H' \leq H$, what is the furthest the ball can land from where it would have landed without deflection?  What are the values of $\theta$ and H'?
Assume the following:  2d world, perfect deflection, zero drag, gravity is G, the ball is a point particle.
 A: Start with the equation for the trajectory of a projectile launched from the origin with speed $u$ at angle $\theta$ to the horizontal, with $\tan\theta=m$ :
$$y=mx-\frac{g}{2u^2}(1+m^2)x^2$$
The deflection point is the origin, at height $h$ above ground. Launch speed is given by $u^2=2g(H-h)$. The projectile hits the ground where $y=-h$.
Let $\frac{u^2}{g}=2(H-h)=a$ and $mx=z$. Substitute into the above equation and rearrange :
$$0=2ah+2az-(x^2+z^2)=(2ah-x^2)+2az-z^2$$
Looking at this as a quadratic in $z=xm$, the range $x$ is a maximum when the two possible values of $z$ are the same, ie when the quadratic is a perfect square. This occurs when $2ah-x^2=-a^2$ giving $z=a$.
The optimum range is given by
$$x^2=a^2+2ah=4(H-h)^2+4(H-h)h=4H(H-h)$$
If $H\ge h \ge 0$ then the maximum range is $x=2H$ when $h=0$. If $h$ can take values below $0$ - ie the deflection point is below ground level - then $x$ can be arbitrarily large.
The optimum angle $\theta$ for a particular value of $h$ can be found from $z=mx=a$ :
$$m=\tan\theta=\frac{a}{x}=\sqrt{\frac{H-h}{H}}$$
Check : When $h=0$ then $\theta=45^{\circ}$. When $h=H$ then $\theta=0^{\circ}$.
The normal to the deflecting plane should make an angle $\frac12(\frac{\pi}{2}-\theta)$ with the vertical.

Another solution starts with a theorem in Mathematics SE question
https://math.stackexchange.com/questions/2660468/projectile-vw-gk-for-minimum-launch-velocity/2687554#2687554.
The minimum launch speed $u$ required to reach a point with horizontal separation $x$ is related to the landing speed $v$ by $$uv=gx$$
Maximum deflection for a given launch speed is equivalent to minimum launch speed to reach a given deflection.
Here $u^2=2g(H-h)$ and $v^2=2gH$ therefore
$$x^2=\frac{u^2v^2}{g^2}=\frac{2g(H-h).2gH}{g^2}=4H(H-h)$$
as found above.
A: Let's set up some variables. By conservation of energy, at height $H'$, the speed is $$v={\sqrt{2g(H-H')}}. \tag{1}$$
The horizontal distance you want to minimize is $$x= v \cos(\theta)t. \tag{2}$$
Next, after deflection, the time it takes to fall is in the following expression:
$$ -H' = -\dfrac{1}{2}gt^2+v\sin(\theta)t \quad\implies\quad H'=\dfrac{1}{2}gt^2-v\sin(\theta)t \tag{3}$$
We combine equations (2) and (3) and get:
$$H'=\dfrac{1}{2}g\dfrac{x^2}{v^2\cos^2\theta} - v\sin(\theta)\dfrac{x}{v\cos\theta}$$
$$H'=\dfrac{gx^2}{2v^2\cos^2\theta} - x\tan(\theta)$$
Then, we combine (3) and finally get:
$$H'=\dfrac{gx^2}{2{\sqrt{2g(H-H')}}^2\cos^2\theta} - x\tan(\theta)$$
$$H'=\dfrac{x^2}{4(H-H')\cos^2\theta} - x\tan(\theta)$$
$$H'\cos^2(\theta)=\dfrac{x^2}{4H-4H'} - \dfrac{1}{2}x\sin(2\theta) \tag{4}$$
Keep in mind that domain for $x$ is $0\leq H' \leq H$ and $0\leq \theta \leq \pi / 2$.
Since $\theta$ and $H'$ are independent of each other, we'd use partial derivatives. First, we'd take partial derivative wrt $\theta$ of the equation in (4), and then set $\partial x / \partial \theta = 0$. That will give us one equation. Then we'd take partial derivative of equation (4) again, only this time wrt $H'$, and set $\partial x / \partial H' = 0$. That will give us second equation. Then solve that system of equations for $h'$ and $\theta$. Interesting problem!
However, since eq (4) is pretty complicated, its partial derivatives will be even more complicated. I would suggest using a software like Maple or Wolfram Alpha to come up with the real solution.
EDIT: made a mistake. fixing it.
A: This document gives an equation for finding the maximum range $R$ of a simple projectile given an initial height from the ground $h$ and velocity $v$:
R = sqrt((2v^2)/g * (h + v^2/(2g)))))
From conservation of energy, we know the velocity $v$ for $H$ and $h$:
v = sqrt(2 * H' * g).
The bigger H', the higher its velocity $v$.
Substituting $v$ into the first equation quite nicely simplifies into:
R = 2 * sqrt(H * H')
For any given $H$ and $H'$, we can now know the optimal range possible.
This appears to be maximal where $H'$ = H. That is, it's best to let the ball fall all the way down.
The document also provides an equation to compute theta given the $h$ and $v$, which we can now produce.  However, it is trivial to see that when $h$ = 0, the optimal angle is 45 degrees.
Note: I'm new here and don't know how to make my equations look nice. I could not find a simple cut and paste thing that will convert simple equations into whatever ridiculous gibberish formats you mathematicians seem content with using :P
