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Solving for initial velocity Projectile Motion

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A rock is thrown from the top of a twenty meter high building at the angle 53 degrees above the horizontal. IF the horizontal range of the throw is equal to the height of the building, what is the speed of the rock? How long is it in the air? What's the velocity of the rock before it hits the ground?

    2. Relevant equations
    Vx = Vox
    Vy = Voy - gt
    x = xo + Voxt
    y = yo + Voy - (1/2)gt^2

    3. The attempt at a solution
    I use the second one and set x = to 20 and solve for the time...

    20 = 0 + Vcos(53)t
    t = 20/Vcos(53)

    I substitute into the y equation

    20 = vsin(53) - 1/2g(20/vcos(53)^2

    Here is where I think I'm making the big mistake... help! :(
  2. jcsd
  3. Feb 24, 2010 #2


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    Homework Helper

    The displacement y is along g. So it must be -ve
    So rewrite the equation and solve for v.
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