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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.

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