Throwing a rock off a cliff: Calculating the optimal angle

You want to throw a rock far, as far as you can possibly throw. Using all your energy, you throw the rock into the sky, and it lands some distance away a few seconds later. But at what angle should you throw it?

It is easy to see that the optimal angle to throw a rock when you are on flat ground is 45^\circ. That is, to make the rock travel as far as possible, one should throw the rock halfway between vertical and horizontal. Any other angle and you do worse than throwing it at a 45^\circ angle.

But what about throwing a rock off a cliff into, say, the ocean? Is 45^\circ still the optimal angle, or is it different? Perhaps it varies?

Suppose you are standing on the edge of a cliff overlooking the sea. It is h meters above the sea, and you can throw a rock with a velocity v, at an angle \theta between 0^\circ and 90^\circ. You are trying to maximize d, the distance thrown.

We assume that there is no friction or air resistance, and we let g be the gravitational constant, 9.81 m/s^2.

When throwing a rock at angle \theta, the velocity of the rock is a vector with a horizontal component of v \cos \theta and a vertical component of v \sin \theta.

Then the distance traveled can be expressed partially in terms of \theta:

d = vt \cos \theta .

The next equation we can get by substituting our values into one of the motion equations:

-h = (v \sin \theta)t - \frac{1}{2}gt^2

The unknown in the two equations is t, or the total time traveled. From the first equation we have

t = \frac{d}{v \cos \theta} .

Substituting into the second equation, we get

-h = (v \sin \theta) (\frac{d}{v \cos \theta}) - \frac{1}{2}g(\frac{d}{v \cos \theta})^2

-h = d \tan \theta - \frac{gd^2 \sec^2 \theta}{2v^2}

From here we want to find the maximum for d, or when \frac{dd}{d \theta}=0. To solve for \frac{dd}{d \theta}, we first differentiate both sides of the equation with respect to \theta:

0 = \frac{d}{d \theta} (d \tan \theta) - \frac{g}{2v^2} \frac{d}{d \theta} (d^2 \sec^2 \theta)

0 = d \sec^2 \theta + \tan \theta \frac{dd}{d \theta} - \frac{g}{2v^2}(\frac{dd^2}{d \theta} \sec^2 \theta + 2d^2 \tan \theta \sec^2 \theta)

As \frac{dd^2}{d \theta} = 2d \frac{dd}{d \theta}, we can isolate \frac{dd}{d \theta}:

\frac{dd}{d \theta} = \frac{\frac{gd^2 \tan \theta \sec^2 \theta}{v^2}-d \sec^2 \theta}{\tan \theta - \frac{gd \sec^2 \theta}{v^2}}

Since \frac{dd}{d \theta} = 0, we eliminate the denominator and are left with

0 = \frac{gd^2 \tan \theta \sec^2 \theta}{v^2} - d \sec^2 \theta

0 = d(\frac{gd \tan \theta \sec^2 \theta}{v^2} - \sec^2 \theta)

Ignoring the trivial case where d=0, we get

gd \tan \theta \sec^2 \theta = v^2 \sec^2 \theta

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

We have the optimal distance; we can substitute this back into a previous equation:

-h = (\frac{v^2}{g \tan \theta}) \tan \theta - \frac{g (\frac{v^2}{g \tan \theta})^2 \sec^2 \theta}{2v^2}

-h = \frac{v^2}{g} - \frac{v^2 \sec^2 \theta}{2g \tan^2 \theta}

-h = \frac{v^2}{g}(1 - \frac{1}{2\sin^2 \theta})

Finally we solve for \theta:

\frac{1}{2 \sin^2 \theta} = \frac{hg}{v^2} + 1

\sin^2 \theta = \frac{v^2}{2hg+2v^2}

\theta = \arcsin(\frac{v}{\sqrt{2hg+2v^2}})

This is the final formula for \theta in terms of v, h, and g.