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Projection motion on a slope; f angles that will provide the greastest range

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A boy is standing on the peak of a hill (downhill), and throws a rock, at what angle from himself to the horizontal should he throw the rock in order for it to travel the greatest distance.
    Answer clues:
    1. if, the angle from the slope to the horizontal = 60, then the angle from the horizontal to the boy =15
    2. the angle is not 45 degrees

    2. Relevant equations

    vf^2 = vo^2 + 2ad
    sin2(theta) = 1
    cos^2(theta) + sin^2(theta) = 1

    3. The attempt at a solution

    I tried to solve this by changing the axis so that the slope is the x axis, and then solving using the first kinematic equation above, i finally ended up solving for vo, which was ( 1+ 2cos(theta) + sin 2(theta) )/2...i doubt that is right though, I just don't know how to put all of this together!
    some help would be appreciated
     
  2. jcsd
  3. Sep 24, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    Hi Sarah, welcome to the Physics Forum!
    Interesting problem, but I can't wrap my mind around this
    This makes no sense to me - could there be a typo?
     
  4. Sep 25, 2009 #3
    no its not a typo..there are two unknown angles, and the clue was that the slope of the hill, or the angle from the slope to the horizontal is 60, then, the angle above that, or the angle between the boy and the horizontal would be 15...i attached a little drawing i made, hope that helps!
     

    Attached Files:

  5. Sep 25, 2009 #4
    I think I solved it. Tough problem. I created 4 equations. The first too were just the x and y positions. The next was for the total distance using the Pythagorean theorem. The final was from realizing that when the object will land, it's ratio of y to x pos will be equal to the tan of the slope (draw a picture).
    After that it just takes a whole lot of manipulation so that you can finally take the derivative and set that equal to zero.
     
  6. Sep 25, 2009 #5
    could you post the four equations you came up with, and i can try to manipulate them myself and see if i can do it?
    thank youu
     
  7. Sep 25, 2009 #6
    [tex] x = v_ocos\theta _2 t [/tex]
    [tex] y = v_osin\theta _2 t -\frac{1}{2}gt^2[/tex]
    [tex]d=\sqrt{x^2+y^2}[/tex]
    [tex]tan\theta _1 = y/x[/tex]
    Manipulation + derivation is a long process. ><
     
  8. Sep 25, 2009 #7
    do you think it would be easier to solve if you tilted the axis so the x axis is parallel to the slope?
     
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