Maximizing Range

The trajectory of the projectile is $x \; = \; vt\cos\theta \hspace{2cm} y \; = \; vt\sin\theta - \tfrac12gt^2$ and so the projectile hits the slope when $y = x\tan\alpha$ , so that $\begin{aligned} y\cos\alpha - x\sin\alpha & = \; 0 \\ vt\sin\theta\cos\alpha - \tfrac12gt^2\cos\alpha - vt\cos\theta\sin\alpha & = \; 0 \\ vt\sin(\theta-\alpha) & = \; \tfrac12gt^2\cos\alpha \end{aligned}$ so that the particle hits the slope when $t = \frac{2v\sin(\theta-\alpha)}{g\cos\alpha}$ . This makes the range of the particle $R(\theta) \; = \; \sec\alpha \times v \times \frac{2v\sin(\theta-\alpha)}{g\cos\alpha} \times \cos\theta \; = \; \frac{2v^2\sin(\theta - \alpha)\cos\theta}{g\cos^2\alpha} \; = \; \frac{v^2}{g\cos^2\alpha}\big[\sin(2\theta -\alpha) - \sin\alpha \big]$ Then $R'(\theta) = \frac{2v^2}{g\cos^2\alpha}cos(2\theta-\alpha)$ and hence $R'(\theta) = 0$ when $2\theta - \alpha = 90^\circ$ , so when $\theta = \tfrac12(\alpha + 90^\circ)$ . It is easy to show that this value of $\theta$ maximizes $R$ .

For this problem, this makes the optimum angle $\tfrac12(20 + 90) = \boxed{55^\circ}$ .

The optimum angle is always halfway between the angle of the slope and the vertical. This principle is still true when firing the projectile downhill.