# Maximum range of a projectile (launched from an elevation)

If a projectile is launched at a speed $u$ from a height $H$ above the horizontal axis, and air resistance is ignored, the maximum range of the projectile is $R_{max}=\frac ug\sqrt{u^2+2gH}$, where $g$ is the acceleration due to gravity.

The angle of projection to achieve $R_{max}$ is $\theta = \arctan \left(\frac u{\sqrt{u^2+2gH}} \right)$.

Can someone help me derive $R_{max}$ as given above?

I have tried substituting $y=0$ and $x=R$ into the trajectory equation

$$y=H+x \tan\theta -x^2\frac g{2u^2}(1+\tan^2\theta),$$

then differentiating with respect to $\theta$ so that we can let $\frac {dR}{d\theta}=0$ (so that $R=R_{max}$), but this would eliminate the $H$, so it won't lead to the expression for $R_{max}$ that I want to derive.

Here's a scan of where I got the problem from:

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 You mean the speed is $u$, not the velocity. That's pretty fundamental. – TonyK Apr 2 '12 at 17:26 Correction made, thanks. – Ryan Apr 2 '12 at 18:31 Cross-posted to physics.stackexchange.com/q/23186/2451 – Qmechanic Apr 2 '12 at 20:54 possible duplicate of What is the optimum angle of projection when throwing a stone off a cliff? – Ross Millikan Apr 2 '12 at 21:52 Problem solved! Please refer to physics.stackexchange.com/a/23187/8446 :-) – Ryan Apr 2 '12 at 23:16

We are given the trajectory of a projectile: $$y = H + x\tan(\theta) - \frac{g}{2u^2}x^2(1+\tan^2(\theta)),$$ where $H$ is the initial height, $g$ is the (positive) gravitational constant and $u$ is the initial speed. Since we are looking for the maximum range we set $y=0$ (i.e. the projectile is on the ground). If we let $L=u^2/g$, then $$H + x\tan(\theta) - \frac1{2L}x^2(1+\tan^2(\theta)) = 0$$ Differentiate both sides with respect to $\theta$. $$\frac{dx}{d\theta}\tan(\theta) + x\;\sec^2(\theta) - \left[\frac1L x \frac{dx}{d\theta}(1+\tan^2(\theta)) + \frac{1}{2L}x^2(2\tan(\theta)\sec^2(\theta))\right] = 0$$ Solving for $\frac{dx}{d\theta}$ yields $$\frac{dx}{d\theta} = \frac{x \sec^2(\theta)[\frac{x}{L}\tan(\theta)-1]}{\tan(\theta)-\frac{x}{L}(1+\tan^2(\theta))}$$ This derivative is $0$ when $\tan(\theta) = \frac{L}{x}$ and hence this corresponds to a critical number $\theta$ for the range of the projectile. We should now show that the $x$ value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace $\tan(\theta)$ with $\frac{L}{x}$ in the second equation from the top and solve for $x$. $$H + L - \frac{1}{2L}x^2 - \frac{L}2 = 0.$$ This leads immediately to $x = \sqrt{L^2 + 2LH}$. The angle $\theta$ can now be found easily.
 Why on earth did you lose $g$? – TonyK Apr 2 '12 at 17:39 I replaced my original answer with this derivation of the maximum range formula. I think this is what you wanted to see. – Patrick Apr 3 '12 at 2:36 @Patrick Thank you, Patrick. The answer I was rather looking for is this exactly: physics.stackexchange.com/questions/23186/… I'm not sure why the source material needed to substitute $L$ for $\frac {u^2}g$... I had un-substituted it back out when posing my question in order to make the expressions cleaner and more useful (I feel). – Ryan Apr 3 '12 at 3:08