What is the optimum angle of projection when throwing a stone off a cliff?

You are standing on a cliff at a height $h$ above the sea. You are capable of throwing a stone with velocity $v$ at any angle $a$ between horizontal and vertical. What is the value of $a$ when the horizontal distance travelled $d$ is at a maximum?

On level ground, when $h$ is zero, it's easy to show that $a$ needs to be midway between horizontal and vertical, and thus $\large\frac{\pi}{4}$ or $45°$. As $h$ increases, however, we can see by heuristic reasoning that $a$ decreases to zero, because you can put more of the velocity into the horizontal component as the height of the cliff begins to make up for the loss in the vertical component. For small negative values of $h$ (throwing up onto a platform), $a$ will actually be greater than $45°$.

Is there a fully-solved, closed-form expression for the value of $a$ when $h$ is not zero?

Solution 1:

Assume no friction and uniform gravity g.

If you throw a stone at point (0, h), with velocity (v cos θ, v sin θ), then we get

\begin{align} d &= vt\cos\theta && (1) \\ 0 &= h + vt\sin\theta - \frac12 gt^2 && (2) \end{align}

The only unknown to be solved is t (total travel time). We could eliminate it by using $t = \frac d{v\cos\theta}$ to get

$$ 0 = h + d\tan\theta - \frac{gd^2\sec^2\theta}{2v^2}\qquad(3) $$

Then we compute the total derivative with respect to θ:

\begin{align} 0 &= \frac d{d\theta}\left(d\tan\theta\right) - \frac g{2v^2}\frac d{d\theta}\left(d^2\sec^2\theta\right) \\ &= \ldots \end{align}

and then set $\frac{dd}{d\theta}=0$ (because it is maximum) to solve d:

$$ d = \frac{v^2}{g\tan\theta} $$

Substitute this back to (3) gives:

\begin{align} h &= \frac{v^2}g \left( \frac1{2\sin^2\theta} - 1\right) \\ \Rightarrow \sin\theta &= \left( 2 \left(\frac{gh}{v^2} + 1\right) \right)^{-1/2} \end{align}

This is the closed form of θ in terms of h.