Maximum range of a projectile (launched from an elevation)

Question

If a projectile is launched at a speed $u$ from a height $H$ above the horizontal axis, $g$ is the acceleration due to gravity, and air resistance is ignored, its trajectory is $$y=H+x \tan θ-x^2\frac g{2u^2}\left(1+\tan ^2\theta\right),$$ and its maximum range is $$R_{\max }=\frac ug\sqrt{u^2+2gH}.$$

I would like to derive the above $R_{\max},$ and here's what I've done:

1. substitute $(x,y)=(R,0)$ into the trajectory equation;
2. differentiate the result with respect to $\theta;$
3. substitute $\left(R,\frac {\mathrm dR}{\mathrm d\theta}\right)=\left(R_{\max},0\right).$

However, this eliminates $H$ and fails to lead to the desired expression for $R_{\max }.$ How to actually derive the above $R_{\max }?$

P.S. This is the context; in the above, I've replaced all occurrences of $L$ below with $\frac{u^2}g$.

Answer 1

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.

Answer 2

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

I would like to derive the above $R_{\max},$ and here's what I've done:

1. substitute $(x,y)=(R,0)$ into the trajectory equation;
2. differentiate the result with respect to $\theta;$
3. substitute $\left(R,\frac {\mathrm dR}{\mathrm d\theta}\right)=\left(R_{\max},0\right).$

This is correct, and gives $$0=R_{\max}\sec^2\theta-R_{\max}^2\frac{g}{2u^2}2\tan\theta\sec^2\theta$$$$\tan\theta=\frac{u^2}{gR_{\max}}.\tag A$$

Step 1 also directly gives $$0=H+R_{\max}\tan θ-R_{\max}^2\frac g{2u^2}\left(1+\tan ^2\theta\right)$$$$\frac {gR_{\max}^2}{2u^2}\left(1+\tan ^2\theta\right)=H+R_{\max}\tan θ.\tag B$$

Substituting $(A)$ into $(B)$ then gives \begin{align}\frac{gR_{\max}^2}{2u^2}\left(1+\frac{u^4}{g^2R_{\max}^2}\right)&=H+\frac{u^2}g\\ \frac{gR_{\max}^2}{2u^2}+\frac{u^2}{2g}&=H+\frac{u^2}g\\ R_{\max}&=\sqrt{\frac{2u^2}g\left(H+\frac{u^2}{2g}\right)}\\ &=\sqrt{\frac{u^2}{g^2}\left(2gH+u^2\right)}\\ &=\frac ug\sqrt{u^2+2gH},\end{align} as desired.