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A 'simple' projectile problem.

  1. Sep 20, 2005 #1
    This problem has been giving me headaches for the past two weeks. Here it is, as well as how far I have gotten on it:

    A projectile is fired with an initial velocity of 53m/s. Find the angle of projection such that the maximum height is equal to it's range.

    So that basically boils down to:

    H = R
    Vo = 53m/s
    ay = -9.81m/s^2
    t = unknown
    x (angle) = find

    According to the nifty sheet the teacher tossed out the applicable formulas would be:

    R=Vo^2*sin(2x)/g (since Ho and Hf will be the same)

    Making H=R gets the formula Voy^2=2*Vo^2*sin(2x) ...which doesn't help me since x is what needs to be found, and without x one can't break Vo into component form.

    Do I have the right idea or have I flubbed it already?


  2. jcsd
  3. Sep 20, 2005 #2
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