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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.

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  • $\begingroup$ Cross-posted from math.stackexchange.com/q/127300/11127 $\endgroup$
    – Qmechanic
    Apr 2 '12 at 20:55
  • $\begingroup$ 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.) $\endgroup$ Apr 2 '12 at 21:23
  • $\begingroup$ H is a constant, so it gets eliminated, no? $\endgroup$
    – Ryan G
    Apr 2 '12 at 21:51
  • $\begingroup$ No. Take $\frac{d}{dx} 5x$. 5 is a constant but does not get eliminated. $\endgroup$ Apr 3 '12 at 0:00
  • $\begingroup$ @Mark, perhaps we're talking at cross-purposes. Please refer to leongz's answer below to see why our H gets eliminated. $\endgroup$
    – Ryan G
    Apr 3 '12 at 0:36
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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 $R_{max}$.

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    $\begingroup$ Beautiful! THANK YOU! ps. I've cleaned up this page to make it more general and useful. $\endgroup$
    – Ryan G
    Apr 2 '12 at 23:54
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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.

Addendum:

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.

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  • $\begingroup$ where did you get that second tan equation from? $\endgroup$ Aug 31 '19 at 18:08
  • $\begingroup$ @Achilles'Advisor - see explanation in Addendum. $\endgroup$ Sep 1 '19 at 18:05

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