# Projectile fired on hill at angle

1. Oct 15, 2005

### eku_girl83

This one is giving me fits!

A projectile is fired with initial speed v0 at an elevation angle of alpha up a hill of slope beta (alpha greater than beta).

How far up the hill will the projectile land?
At waht angle alpha will the range be a maximum?
What is the maximum range?

Where do I begin? How do I write (or alternately, derive) the equations of motions for this case? Normally, for a projectile fired at an angle, we divide the velocity into components and then apply equations for two-dimensional motion. But I just don't know how to approach a projectile at an angle which is fired at another angle!

Any help greatly appreciated!!

2. Oct 15, 2005

### Skippy

When I read it, I was thinking the place from where you fire the projectile from is flat and then there's a hill in front of it. (ie alpha and beta both measured from the x-axis)

Sounds like you're saying alpha is the angle between the projectile and the hill, if that's the case then you just add the two angles to get the angle at which the projectile is fired (relative to the x-axis). In that case though, why would it say "alpha greater than beta".

Either way you choose to use the angle... you can still get your *time when ball hits hill* using the Y motion of the projectile, you'll just have an extra term in there that'll relate how the height of the ground changes over time.

Last edited: Oct 15, 2005
3. Oct 15, 2005

### pymike

That does seem like a tough one. I think you could do it numerically if you were given values but it sounds like you are wanted to do it algebracially. I'm a bit bit stumped by it myself, this might help but probably not . If I have the problem set up right, I believe you can set it up for the tan(beta), then have that equal to the equation for x over the equation for y. I'm not sure what to do to maximize x for alpha though. Ok, hope I can get this TeX right...

$$\tan\beta = \frac{x}{y}$$

$$x = V_{xo}t = \cos({\alpha})V_{o}t$$

$$y = V_{yo}t + \frac{1}{2}gt^2 = \sin({\alpha})V_{o}t + \frac{1}{2}gt^2$$

Last edited: Oct 15, 2005
4. Oct 17, 2005

### hotvette

I believe this situation boils down to finding the intersection of 2 functions. One function is the y = f(x) to describe the projectile as if there were no incline. The second function describes the incline, which is just a line. Find the intersection of the two functions.