Firing a Cannon

Let the initial velocity of the ball be $u$ . Then by the conservation of energy $\color{#3D99F6} \frac 12 mu^2 = mgL$ . Now consider the ball moving on the inclined plane. Let the velocity when the ball emerges from the plane be $v$ . Then by the conservation of energy again, we have:

$\begin{aligned} {\color{#3D99F6} mgL}\sin \theta + \frac 12 mv^2 & = \frac 12 mu^2 & \small \color{#3D99F6} \text{Note that }\frac 12 mu^2 = mgL \\ {\color{#3D99F6}\frac 12 mu^2} \sin \theta + \frac 12 mv^2 & = \frac 12 mu^2 \end{aligned}$

$\implies v^2 = u^2(1-\sin \theta)$

Now we know that the range $d$ of the trajectory is given by:

$\begin{aligned} d & = \frac {\color{#3D99F6}v^2}g \sin (2\theta) & \small \color{#3D99F6} \text{Since }v^2 = u^2(1-\sin\theta) \\ & = \frac {\color{#3D99F6}u^2}g {\color{#3D99F6}(1-\sin \theta)}\sin (2\theta) \end{aligned}$

We note that $d$ is maximum when $f(\theta) = (1-\sin \theta) \sin (2\theta)$ is maximum.

$\begin{aligned} f'(\theta) & = 2(1-\sin \theta) \cos (2\theta) - \cos \theta \sin (2\theta) \\ & = 2(1-\sin \theta)(1-2\sin^2 \theta) - 2 \sin \theta (1-\sin^2 \theta) \\ & = 2(1-\sin \theta)(1-2\sin^2 \theta - \sin \theta - \sin^2 \theta) \\ & = 2(1-\sin \theta)(1-\sin \theta - 3\sin^2 \theta) \end{aligned}$

Putting $f'(\theta)=0$ to find the extrema, we have

$\begin{cases} \sin \theta = 1 \implies \theta = \dfrac \pi 2 & \text{when } d = 0 \text{, the minimum} \\ \sin \theta = x = \dfrac {\sqrt {13}-1}6 \approx \boxed{0.434} & \text{when } d \text{ is maximum} \end{cases}$