1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Projectile up a slope

  1. Oct 5, 2007 #1
    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.

  2. jcsd
  3. Oct 5, 2007 #2
    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.
  4. Oct 6, 2007 #3
    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.

    [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.
  5. Oct 6, 2007 #4
    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.
  6. Oct 6, 2007 #5
    So when you get R. Try to find out that for which angle the R would be maximum.
  7. Oct 6, 2007 #6
    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.
  8. Oct 6, 2007 #7
    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

    [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.If not i am always here.
    Last edited: Oct 6, 2007
  9. Oct 7, 2007 #8
    I have figured it out. Thanks a bunch.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook