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Projectile Motion (find theta)

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data

    The speed of a projectile when it reaches its maximum height is 0.31 times 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 θ) / 2g

    R=(Vi^2 * sin2θ) / g

    3. The attempt at a solution

    I realize that i need to somehow come up with a ratio and set the two equal to each other. However, i'm having difficulty with the initial step. Maybe I could get a hint that could set me on the right track?
     
  2. jcsd
  3. Sep 14, 2013 #2
    Decompose the velocity vector.
    At the top which component of velocity is zero?
    Use this idea to find the ratio. As you mentioned
     
  4. Sep 14, 2013 #3
    So after decomposing the velocity vector I got:

    Vi = -yf-at2 /t

    Plugging that in to the equation for h...

    h= ((-yf-at2)2*(sin2θ)) / 2g

    and then

    0.5h = 0.31((-yf-at2)2*(sin2θ)) / 2g

    Am I on the right track?
     
  5. Sep 14, 2013 #4
    You see at the top point V is just Vx so can you relate this to the given information.
     
  6. Sep 15, 2013 #5
    I'm sorry for late response I got lectures :)
    So you still solving the problem?
     
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