What is the relationship between launch height and the launch angle for maximum range? My physics teacher gave this question to our class and it has me stumped. I cant find any written examples on the internet talking about the relationship between an elevated projectile and the optimum angle for the furthest range. was hoping someone might be able to help me out. Thanks 
 A: Have you tried to search the net for 'projectile initial height'?
A few first hits contain e.g.


*

*Range of a projectile article in Wikipedia,

*Solving for the initial height of a projectile in Physics Stack Exchange, and

*Projectile Motion Range, Initial Height, and Maximum height in YouTube,


which all seem to give answer to your problem.
EDIT
Also Maximum range of a projectile (launched from an elevation) here, at Math SE, seems to be the same problem as yours.
A: Assume projectile is launched from height of $h$. 
Let $v=$launch velocity, and $g=$gravity. 
Trajectory equation for projectile is
$$y-h=x\tan\theta-\frac {gx^2}{2v^2}\sec^2\theta$$
Putting $t=\tan\theta$, and $k=\displaystyle\frac g{2v^2}$, equation becomes
$$y-h=tx-k(t^2+1)x^2$$
Let $r=$horizontal range when projectile hits the ground, i.e. at $y=0$.
$$-h=tr-k(t^2+1)r^2\\
k(t^2+1)r^2-tr-h=0\qquad\qquad(1)$$
Differentiating w.r.t. t, 
$$k(t^2+1)2r\frac {dr}{dt}+k(2t)r^2-t\frac{dr}{dt}-r=0$$
At maximum $r$, $\displaystyle\frac{dr}{dt}=0$
$$r(1-2ktr)=0\\
\because r\neq0 \therefore r=\frac 1{2kt}$$
Substituting in (1) and rarranging gives
$$t=\frac 1{\sqrt{1+4kh}}\\
\color{red}{\tan\theta^*=\frac 1{\sqrt{1+\frac {2gh}{v^2}}}}$$
This is the angle (or rather its tangent) which gives maximum range. 
The maximum range is given by 
$$r^*=\frac {\sqrt{1+4kh}}{2k}\\
r^*=\frac {v^2}g\sqrt{1+\frac {2gh}{v^2}}=\color{orange}{\frac vg\sqrt{v^2+2gh}}$$
See illustration here.

Addendum (added 29 March 2018)
See also the question and solutions here.
The problem here is to find the optimal angle for the furthest range at a fixed launch velocity. Thisis the inverse of the problem in question $2660468$, which is the find the minimum velocity (and corresponding optimal angle) for a fixed range. 
Also in the problem here the launch elevation is higher than the destination, while this is the opposition in question $2660468$. To make this equivalent we can simply assume that the projectile in question $2660468$ is launched in reverse, i.e. at the terminal velocity and angle. 
Using the results in question $2660468$ (and using notation in this question), note that
$$v^2=g(R-h)\\w^2=g(R+h)$$
Optimal angle $\phi$ is given by 
$$\tan\phi=\frac {R-h}k=\frac {R-h}{\sqrt{R^2-h^2}}=\frac {\frac {v^2}g}{\frac vg\sqrt{v^2+2gh}}=\color{red}{\frac v{\sqrt{v^2+2gh}}=\frac 1{\sqrt{1+\frac {2gh}{v^2}}}}$$
Maximum range $r^*$ is given by 
$$r^*=k=\frac {vw}g=\color{orange}{\frac {v{\sqrt{v^2+2gh}}}g}$$
