Problem source

A cannon-ball is fired from a cannon at an initial speed $u$. After time $t$ it has reached height $h$ and is at a distance $\sqrt{x^{2}+h^{2}}$ from the cannon. Ignoring air resistance, show that
\[
\tfrac{1}{4}g^{2}t^{4}-(u^{2}-gh)t^{2}+h^{2}+x^{2}=0.
\]
Hence show that if $u^{2}>2gh$ then the horizontal range for a given height $h$ and initial speed $u$ is less than or equal to 
\[
\frac{u\sqrt{u^{2}-2gh}}{g}.
\]
Show that there is always an angle of firing for which this value is attained.

Solution source

Suppose it is fired with angle to the horizontal $\alpha$, then
\begin{align*}
\rightarrow: && x &= u\cos \alpha \cdot t \\
\uparrow: && h &= u \sin \alpha \cdot t - \frac12 g t^2 \\
\Rightarrow && u\cos \alpha &= \frac{x}{t} \\
&& u \sin \alpha &= \frac{h + \frac12 gt^2}{t} \\
\Rightarrow && u^2 &= \frac{x^2}{t^2} + \frac{(h + \frac12 gt^2)^2}{t^2} \\
\Rightarrow && 0 &= x^2+h^2-u^2t^2+ght^2+\tfrac14 g^2 t^4 \\
&&&= \tfrac14 g^2 t^4 - (u^2 - gh)t^2 + h^2 + x^2
\end{align*}

For a distance $x$ to be achievable there must be a root to this quadratic in $t^2$, ie
\begin{align*}
&& 0 &\leq \Delta = (u^2-gh)^2 - 4 \cdot \tfrac14 g^2 (h^2 + x^2) \\
\Rightarrow && x^2 &\leq \frac{(u^2-gh)^2}{g^2} - h^2 \\
&&&= \frac{u^4+g^2h^2 - 2ghu^2-g^2h^2}{g^2} \\
&&&= \frac{u^2(u^2-2gh)}{g^2} \\
\Rightarrow && x &\leq \frac{u\sqrt{u^2-2gh}}{g}
\end{align*}
This is achieved when  
\begin{align*}
&& t^2 &= \frac{u^2-gh}{\tfrac12g^2}\\
&&&= \frac{2(u^2-gh)}{g^2} \\
\Rightarrow && \cos \alpha &= \frac{u\sqrt{u^2-2gh}}{g} \cdot \frac{g}{\sqrt{2(u^2-gh)}} \frac{1}{u} \\
&&&= \frac{1}{\sqrt{2}}
\end{align*}
ie when $\alpha = \frac{\pi}{4}$