# Maximize the distance of projectile on a hill

1. Sep 6, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
An athlete stands at the peak of a hill that slopes downward uniformly at angle $\phi$. At what angle $\theta$ from the horizontal should they throw a rock so that it has the greatest range?

2. Relevant equations
SUVAT equations and trig relations

3. The attempt at a solution

I found the following three equations:

$y = v_0 \sin \theta t - \frac{1}{2} g t^2$
$t = \frac{x}{v_0 \cos \theta}$
$y = -x \tan \phi$

I combined all of these to find the final horizontal position x:

$x = \frac{v_0^2}{g}(\sin 2 \theta + 2 \tan \phi \cos^2 \theta)$
We need to maximize $x$ with respect to $\theta$, so we take the first derivative and set it to $0$.

$\frac{dx}{d \theta} = 2 \cos 2 \theta + 2 \tan \phi (2 \cos \theta) (- \sin \theta) = 0$

$\tan 2 \theta = \frac{1}{ \tan \phi}$

So this is what I end up with, but I am not sure what to do next... Any ideas?

Edit: I played with it a little more and found that $\theta = \frac{1}{2} (90 - \phi)$, but I am not quite sure how to use this to find the angle that gives the maximum distance.

2. Sep 6, 2016

### ehild

You know tan(2θ). What is θ then? (Imagine a numerical value for tan(Φ). How would you calculate θ? )

3. Sep 6, 2016

### Simon Bridge

Your analysis shows that the angle to get maximum distance depends on the slope ... if you expect the max distance angle to be independent of the slope, then you have made a mistake somewhere.

Note: what do they mean by "range"?

4. Sep 6, 2016

### PeroK

That was an excellent solution. And that is the answer. You have found $\theta$ in terms of $\phi$.

What you have is called a general solution to the problem: you have established a direct relationship between the two variables.

Now, if you know or measure $\phi$ for a given slope, you can calculate the required $\theta$. You could then calculate the range, $x$, using these values.

Alternatively, you could go back to your equation:

$x = \frac{v_0^2}{g}(\sin 2 \theta + 2 \tan \phi \cos^2 \theta)$

And, using some trig, express $x$ entirely as a function of $\phi$ using $\theta = \frac{1}{2} (90 - \phi)$

That would give you a general solution for the maximum possible range, $x$, in terms of $\phi$.

5. Sep 6, 2016

### rude man

Yes, I have an idea - you're done, essentially!
To finish up,
tan(2θ) = cot(φ)
2θ = tan-1[cot(φ)]
& I'll leave the rest to you! :-)
Well done.

EDIT: and yes I see that θ = 1/2 (90 - φ) is equivalent and better.

6. Sep 6, 2016

### Mr Davis 97

Awesome! I'll try to remember that in the future, that the solution might just be a symbolic relationship rather than numerical.