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Projection angle using max. height w/ proportions only

  1. Sep 19, 2008 #1
    Figures that the only problem I have trouble with is the one the book considers to be "easy":

    1. The problem statement, all variables and given/known data
    The speed of a projectile when it reaches its maximum height is one half its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

    2. Relevant equations
    h= (vi^2 sin^2 theta)/2g

    3. The attempt at a solution

    Complete brain fart--I don't even know where to start, beyond trying to divide h/2, which would be (vi^2 sin^2 theta)/4g, correct? That's about as far as I got. I just keep miring myself the further into it I go.
  2. jcsd
  3. Sep 20, 2008 #2


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    Staff Emeritus
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    Gold Member

    Speed of course is the magnitude of velocity. At its maximum height a projectile only has one component of velocity. Is that enough to get you started?
  4. Sep 20, 2008 #3
    Try breaking velocity into x and y components @ 1/2 max height and then finding speed from there the "sqrt of the sum of the squares".
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