Problem source

A child is playing with a toy cannon on the floor of a long railway carriage. The carriage is moving horizontally in a northerly direction with acceleration $a$. The child points the cannon southward at an angle $\theta$ to the horizontal and fires a toy shell which leaves the cannon at speed $V$.  Find, in terms of $a$ and $g$, the value of $\tan 2\theta$ for which the cannon has maximum range (in the carriage). 
If $a$ is small compared with $g$, show that the value of $\theta$ which gives the maximum range is approximately  
\[
\frac \pi 4 + \frac a {2g},
\]
and show that the maximum range is approximately $\displaystyle \frac {V^2} g + \frac {V^2a}{g^2}. $

Solution source

\begin{align*}
&& s_x &= V \cos \theta t + \frac12at^2 \\
&& s_y &= V \sin \theta t - \frac12 gt^2 \\
\Rightarrow && T &= \frac{2V \sin \theta}g \\
\Rightarrow && s_{max} &= \frac{2V^2 \sin \theta \cos \theta}{g} + \frac12a \frac{4V^2 \sin^2 \theta}{g^2} \\
&&&= (g \sin 2 \theta+2a\sin^2 \theta)\frac{V^2}{g^2} \\
&& \frac{\d s_{max}}{\d \theta} &= (2g \cos 2 \theta +4 a \cos \theta \sin \theta)\frac{V^2}{g^2} \\
&&&= (2g \cos 2\theta + 2a \sin2 \theta) \frac{V^2}{g^2} \\
\Rightarrow && \tan 2\theta &= -\frac{a}{g} \\
\Rightarrow && 2 \theta &\in (\frac{\pi}2, \pi) \\
\Rightarrow && \tan \left (\frac{\pi}{2} - 2 \theta\right) &=-\frac{a}{g} \\
\Rightarrow && \frac{\pi}{2} - 2 \theta&\approx  -\frac{a}{g} \\
\Rightarrow && \theta &\approx \frac{\pi}{4} + \frac{a}{2g} \\
\\
&& s_{max} & \approx  \left (g \sin \left (\frac{\pi}{2} + \frac{a}{g} \right)+2a\sin^2 \left ( \frac{\pi}{4} + \frac{a}{2g}\right)\right)\frac{V^2}{g^2} \\
&&&\approx \left (g \cdot 1+2a\left( \frac{1}{\sqrt{2}}(\frac{a}{2g}+1)\right)^2\right)\frac{V^2}{g^2} \\
&&&\approx \left (g+a\left(1+\frac{a}{g}\right)^2\right)\frac{V^2}{g^2} \\
&&&\approx \left (g+a\right)\frac{V^2}{g^2} \\
&&&= \frac{V^2}{g} + \frac{V^2a}{g}
\end{align*}