# 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 $$θ = \arctan \left(\frac u{\sqrt{u^2+2gH}} \right)$$.

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

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

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

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: 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

There. is actually a much "classier, old school solution" to this problem. I came across it as a question in an older A level M2 textbook by a remarkably inventive author D. Quadling . I confess that I used the brute force differentiation method to get through it but realised that Quadling had laid out the problem with a strong hint that a rather lovely simple solution was possible without calculus. Here are some hints, do shout if you need more( one needs to unravel this oneself to really appreciate its elegance.

Take the standard cartesian trajectory equation but with the emuzzle speed, u,written as u= sqrt(ag).Now complete the square eg form ( xtan(theta)-a)^2 etc Iam sure you will agree the result is stunning .

Chris

Adapting concepts from the question and solutions here, we have

$$R_\text{max}=\frac {uw}g=\frac {u\sqrt{u^2+2gH}}g=\color{red}{\frac ug\sqrt{u^2+2gH}}$$ and $$\tan\phi=\frac {\ell-H}{\sqrt{\ell^2-H^2}} =\frac {\frac {u^2}g}{\frac ug \sqrt{u^2+2gH}}=\color{red}{\frac u{\sqrt{u^2+2gH}}}$$ where $\ell$ is the linear distance between the launch and end points of the projectile.