Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Cross-posted from – Qmechanic Apr 2 '12 at 20:55
Differentiating will not eliminate H. You need the derivative of x with respect to $\theta$. I can read off from what you have that $H$ is divided by $\tan\theta$ when you solve for $x$. (I didn't check everything else, though.) – Mark Eichenlaub Apr 2 '12 at 21:23
H is a constant, so it gets eliminated, no? – Ryan Apr 2 '12 at 21:51
No. Take $\frac{d}{dx} 5x$. 5 is a constant but does not get eliminated. – Mark Eichenlaub Apr 3 '12 at 0:00
@Mark, perhaps we're talking at cross-purposes. Please refer to leongz's answer below to see why our H gets eliminated. – Ryan Apr 3 '12 at 0:36
up vote 2 down vote accepted

As you described, we substitute $y=0$ and $x=R$ into the trajectory equation: $$0=H+R\tan{\theta}-R^2\frac{g}{2u^2}\sec^2\theta.\tag{1}$$ Then, differentiating with respect to $\theta$ and setting $\frac{dR}{d\theta}=0$:

$$0=R_{max}\sec^2\theta-R_{max}^2\frac{g}{2u^2}2\sec^2\theta\tan\theta,$$ which simplifies to $$R_{max}=\frac{u^2}{g}\cot\theta.\tag{2}$$ Solving $(1)$ and $(2)$ will yield the desired expressions for $\theta$ and $R_{max}$.

share|cite|improve this answer
Beautiful! THANK YOU! ps. I've cleaned up this page to make it more general and useful. – Ryan Apr 2 '12 at 23:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.