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Projectile, max height half the range, find angle

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data

    find the angle of the projection for which the maximum height is equal to half of the range

    2. Relevant equations
    vertical: h =(vsinθ)t-0.5(g)(t^2)
    horizontal: 2h =(vcosθ)t



    3. The attempt at a solution

    i know you set both equations equal to each other
    (vsinθ)t -(0.5)(g)(t^2) = ((t)vcosθ)/2

    cancel t
    (vsinθ)-(0.5)(g)(t) = (vcosθ)/2

    (2vsinθ-vcosθ)/2 = (0.5)(g)(t)
    at this point i am slightly stuck, i know i need to sub some equivalent form in for "t" though i am not sure how/what formulas to use.
     
  2. jcsd
  3. Sep 27, 2012 #2

    PeterO

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    Homework Helper

    Go for concepts here.

    At maximum height, the object is half way (horizontally) to its landing point - so at that point the height is 1/4 of the range.

    The vertical part of the trip has been done while slowing down, so the average speed is 1/2 the initial speed.

    It gets to a height 4 times its range so far, so the average vertical speed is 4x the horizontal speed.

    Thus the initial vertical speed is 8x the horizontal speed.

    Apply Pythagoras to that.
     
  4. Sep 27, 2012 #3
    I was able to figure it out but thanks!!
     
  5. Sep 27, 2012 #4

    PeterO

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    Re-reading to see that you were only after the angle, you merely wanted tan-1(8)
     
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