Maximizing Range 2

The trajectory of the particle is $x \; = \; vt\cos\theta \hspace{2cm} y \; = \; h + vt\sin\theta - \tfrac12gt^2$ or $y \; = \; h + x\tan\theta - \tfrac{g}{2v^2}x^2\sec^2\theta$ and so the particle hits the ground when $h + x\tan\theta - \tfrac{g}{2v^2}x^2\sec^2\theta = 0$ . Thus the range $x$ is maximized when $x'(\theta) = 0$ , and we see that $\begin{aligned} x' \tan\theta + x \sec^2\theta - \tfrac{g}{v^2}xx'\sec^2\theta - \tfrac{g}{v^2}x^2\sec^2\theta\tan\theta & = \; 0 \\ \big(\tan\theta - \tfrac{g}{v^2}x\sec^2\theta\big)x' & = \; x\sec^2\theta\big(\tfrac{g}{v^2}x\tan\theta - 1\big) \end{aligned}$ and so maximum range is achieved when $\tfrac{g}{v^2}x\tan\theta = 1$ , and hence $\begin{aligned} h + \tfrac{v^2}{g} - \tfrac{g}{2v^2}\big(x^2 + \tfrac{v^4}{g^2}\big) & = \; 0 \\ x^2 & = \; \tfrac{v^2}{g^2}(v^2 + 2gh) \end{aligned}$ Thus the maximum range is $\tfrac{v}{g}\sqrt{v^2 + 2gh}$ , achieved when $\tan\theta = \tfrac{v}{\sqrt{v^2 + 2gh}}$ . In this case, with $v^2 = gh$ , we have $\tan\theta = \tfrac{1}{\sqrt{3}}$ , and so $\theta = \boxed{30^\circ}$ .

A solution without derivation.