Maximizing Area with a Projectile

At any time $t$ , the projectile's horizontal and vertical displacement are

$\begin{aligned} x&=vt\cos\theta,\\ y&=vt\sin\theta-\frac 12gt^2. \end{aligned}$

If $t$ is eliminated between these two equtions, the following eqution is obtained

$y=x\tan\theta-\frac g{2v^2\cos^2\theta}x^2.$

The projectile hits the slope when

$y=x\tan\alpha.$

Substituting this value into the equation above, we get

$x(\tan\theta-\tan\alpha)=\frac g{2v^2\cos^2\theta}x^2.$

Solving for $x$ , we have

$x=0,\ x=\frac {2v^2\sin(\theta-\alpha)\cos\theta}{g\cos\alpha}.$

So the area between the trajectory and the inclined plane is

$\begin{aligned} A(\theta)&=\int_0^{2v^2\sin(\theta-\alpha)\cos\theta/g\cos\alpha}\left[\left(x\tan\theta-\frac g{2v^2\cos^2\theta}x^2\right)-x\tan\alpha\right]dx \\&=\left.\left[\frac 12x^2(\tan\theta-\tan\alpha)-\frac g{6v^2\cos^2\theta}x^3\right]\right|_0^{2v^2\sin(\theta-\alpha)\cos\theta/g\cos\alpha} \\&=\frac {2v^4}{3g^2\cos^3\alpha}\sin^3(\theta-\alpha)\cos\theta. \end{aligned}$

Setting $A'(\theta)$ equal to zero gives

$\begin{aligned} 0&=3\cos(\theta-\alpha)\cos\theta-\sin(\theta-\alpha)\sin\theta \\\implies\qquad0&=3-\tan(\theta-\alpha)\tan\theta \\\implies\qquad0&=3(1+\tan\theta\tan\alpha)-(\tan\theta-\tan\alpha)\tan\theta \\\implies\qquad0&=\tan^2\theta-4\tan\alpha\tan\theta-3. \end{aligned}$

The physical root of the quadratic equation with respect to $\tan\theta$ is

$\tan\theta=2\tan\alpha+\sqrt{4\tan^2\alpha+3}.$

It is easy to show that this value of $\theta$ maximizes $A$ .

In this case, with $\tan\alpha=\sqrt 3-1$ , we have $\tan\theta=2+\sqrt 3$ , and so $\theta=\boxed{75^\circ}$ .