# Projectile up a slope

1. Oct 5, 2007

### FissionMan1

How do you find the maximum range a projectile will hit (shot at angle alpha) on a slope (angle theta) when theta is less than alpha. Also, what is the best angle for alpha. I would only like a head start, do not post the answer please.

We know that there are constants alpha, theta, and Vo.

I derived the equation for the distance (L) the projectile will travel up the slope.

L=(2Vo^2)/g*(cos(a)/cos(b))*(sin(a)-cos(a)tan(b))

2. Oct 5, 2007

### FissionMan1

My thoughts are that we make beta = 0 and alpha = 45, but I believe the professor is looking for an equation instead of an actual number so that I can prove it is the best angle.

3. Oct 6, 2007

### FedEx

Here we will have to develop a good relation.

Now the angle of the plane is theta and the angle of the projection is theta.
Now let the ground be represented by a line,AB. And the plane is AC

And the angle between AB and AC is alpha. Now you will have to use the following ideas.

$$1) cos\alpha\ = \frac{AB}{AC}$$

2) The vertical displacement is zero wrt AC so now you can apply the equation of kinematics which covers time,initial velocity in y direction and g.
Now we can get time out of it.Now we know that the horizontal comp of velocity is constant and hence we can say that

$$AB = ucos\{\alpha\ + \theta\}T$$

And now use 1 so you will get AB. Which is the range and now by using logic and trigonometry you can find the angle for which R is max.

4. Oct 6, 2007

### FissionMan1

I understand how to find R, but I'm not trying to find that. I am attempting to find the angle at which the projectile will land the farthest up the hill for a given Vo.

5. Oct 6, 2007

### FedEx

So when you get R. Try to find out that for which angle the R would be maximum.

6. Oct 6, 2007

### FissionMan1

R, according to your words, is the projectile's distance over a flat surface. I am trying to find the angle of maximum projection up a slope as a function of AC and theta.

7. Oct 6, 2007

### FedEx

Hold on. What are you trying to say.

I will have to break the forum rules as you are not understanding what i am trying to say.

The equation for AC is as follows

$$\frac{2u^2sin(\phi\ - \theta)cos\phi}{gcos^2\theta}$$

where phi is alpha + theta. Now simplify it and you will get the following for AC

$$\frac{2u^2sin[(2\phi\ - \theta) - sin\theta]}{gcos^2\theta}$$

To get maximum AC here

$$sin(2\phi\ - \theta) = 1$$

Hence in that case max range would be

$$\frac{u^2}{g(1 + sin\theta)}$$

All this equations, I have derived by the method i gave you in the last post.

And now from the above three equations you can find out the angle.If not i am always here.

Last edited: Oct 6, 2007
8. Oct 7, 2007

### FissionMan1

I have figured it out. Thanks a bunch.