# 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 $g$.

## 7 thoughts on “Throwing a rock off a cliff: Calculating the optimal angle”

1. bbzippo says:

Hi lucky,
I recall solving this problem a long-long time ago…
Yes, it requires calculus, unlike the 45 degree problem which i wrote about here http://bbzippo.wordpress.com/2010/01/14/45-degrees-proof/ and here http://bbzippo.wordpress.com/2010/01/19/45-degrees-optimum-geometric-proof/

But as far as I recall, I applied some “symmetry” argument to simplify the derivation.
I think the idea was to extend the parabola beyond the point from which we are shooting. I believe that you can show that the _landing_ angle of the optimal trajectory must still be 45 degrees. Is this correct? Would this lead to a shorter derivation?
Sorry, I only have residual memories, and no time for serious thinking…

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2. luckytoilet says:

I don’t think that’s correct. Simply plugging in arbitrary values (g=9.81m/s^2, h=100m, v=30m/s) we get $\theta = 29.282^\circ$, and d=163.601m. Solving the quadratic, we get $y=-0.0072x^2+0.561x+100$ which lands at an angle of about 31 degrees.

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1. parityerror says:

31 Degrees ? or is it about 61 Degrees.

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1. luckytoilet says:

Oh yes it’s 61 degrees.

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3. Nikola Kostic says:

The issue here is you solve for optimal theta with v as a variable. But if you throw an object on a flat survace (delta h =0) the optimal angle is going to be 45 regardless of velocity. This equation should also be independent of initial velocity, basically optimal angle as a function of only the difference in height (assuming g is constant)

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4. Bob says:

Thank you so much for this! I spent 4 hours trying to figure out how to solve this problem yesterday before I finally found your guide.
The way I had been approaching it was to eliminate t for the distance equation and multiply that by the final time. But it left me with this incredibly difficult trig equation that I couldn’t simplify even after several hours.

What I am now wondering is whether there’s a general principle that would have allowed me to determine beforehand how to approach this problem.

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