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

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.

• Cross-posted from math.stackexchange.com/q/127300/11127 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.) Apr 2 '12 at 21:23
• H is a constant, so it gets eliminated, no? Apr 2 '12 at 21:51
• No. Take $\frac{d}{dx} 5x$. 5 is a constant but does not get eliminated. 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. Apr 3 '12 at 0:36

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 $R_{max}$.

• Beautiful! THANK YOU! ps. I've cleaned up this page to make it more general and useful. Apr 2 '12 at 23:54

Adapting concepts from the question and solutions here, we have

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

Here $$\tan\theta^*$$ was derived using $$\theta^*=\frac \pi 4+\frac\alpha 2$$, which leads to $$\tan\theta^*=\frac {1+\sin\alpha}{\cos\alpha}=\frac {\ell (1+\sin\alpha)}{\ell\cos\alpha}=\frac{\ell -H}{R_{\text{max}}}$$ as $$\alpha$$ is negative in this case.