Projectile Motion on an Incline

  • #1

Homework Statement

A baseball is given an initial velocity with magnitude [tex]v_{0}[/tex] at an angle of [tex]\phi[/tex] above the surface of an incline, which is in turn inclined at an angle [tex]\theta[/tex] above the horizontal.

a) Calculate the distance measured along the incline from the launch point to where the baseball strikes the incline. In terms of [tex]v_{0}, g, \theta, \phi[/tex].

b) What angle [tex]\phi[/tex] gives the maximum range, measured along the incline.

Homework Equations

[tex]x_b = v_{0}\cos{(\theta+\phi)}t[/tex]
[tex]y_b = v_{0}\sin{(\theta+\phi)}t - \frac{1}{2}gt^2[/tex]

[tex]y_i = x\tan{\theta}[/tex]

The Attempt at a Solution

Part a) is fairly straightforward to solve... eliminating t to find [tex]y(x)[/tex] and letting [tex]y_i = y_b[/tex] yields:

[tex]d = \frac{2v_{0}^2}{g}\frac{\cos^2{(\theta+\phi)}}{\cos{\theta}}\left[\tan{(\theta+\phi)}-\tan{\theta}\right][/tex]

Which is the solution to a). Just having trouble finding a worked solution for part b). My maths is a little shaky, can someone walk me through it? It's easy enough to find a partial derivative for d, but I still can't solve it for 0.

Answers and Replies

  • #2
Hmm. So you decided to go with the method of a constraint? I used a rotated coordinate system and got something a little a different. But if you say that the solution you got to a) is correct, then I'll just let it go.

As far as finding the maximum angle would go. It sounds like you took the partial of d with respect to phi, though if theta and vo are fixed as the problem says then in actuality this would be a full derivative as d would not depend on theta or vo. Regardless, you are right, the derivative of what you have is very hard to solve for when you set it to zero. I can't do it, and Mathematica (a math program) can't do it. If you are given theta then you could do it numerically.
  • #3
Yeah, the solution is quoted in the back of the book, pt a) is correct. The solution to pt b) is [tex]\frac{\pi}{4}-\frac{\theta}{2}[/tex]. I'd like to know how it was solved though. Since the expression for d I found is the solution quoted for pt a) by the text, it seems logical to assume that the solution for b) follows on.
  • #4
Let us consider a general case

Lets say that a projectile is shot at alpha on an incline of angle theta.

Here we will have to develop a good relation.

Now the angle of the plane is theta and the angle of the projection is alpha.
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.

[tex]1) cos\alpha\ = \frac{AB}{AC}[/tex]

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

[tex] AB = ucos\{\alpha\ + \theta\}T[/tex]

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.

The equation for AC is as follows

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

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

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

To get maximum AC here

[tex]sin(2\phi\ - \theta) = 1[/tex]

Hence in that case max range would be

[tex]\frac{u^2}{g(1 + sin\theta)}[/tex]

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.
  • #5
I can't follow that at all..

Firstly shouldn't it be:


The next equation is just the calculus-derived expression for x, but I don't see how you're getting to the next equation, nor how you're "simplifying" to get the proceeding one..
  • #6
I just reworked the problem using the plane of the incline as the x-axis and managed to find for x:

[tex]x = \frac{2v_{0}^2\cos^2{\phi}\left[\tan{\theta}+\tan{\phi}\right]}{g}[/tex]

Which I assume is what Mindscrape found...

Which makes:

[tex]d = \frac{2v_{0}^2\cos^2{\phi}\left[\tan{\theta}+\tan{\phi}\right]}{g\cos{\theta}}[/tex]

Where d is the distance along the incline rather than across the axis. This expression is more easily differentiable than the one I obtained via the first method... anyway a little algebra and an arctan result I had to search on wikipedia for (never used it before), managed to get the solution:

[tex]2\phi = \arctan{\left(\frac{1}{\tan{\theta}}\right)}[/tex]
[tex]\phi = \frac{\pi}{4}-\frac{\theta}{2}[/tex]

Anyone have any ideas on how this could be solved using the first method? I'm sure it's possible, since the value of d quoted as the solution by the text is as worked via the first method...
  • #8
Looks good to me that's what our TA got in class tonight.

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