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Projectile motion

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data
    a tennis player standing 12.3 m from the net hits the ball at 3° above the horizontal. to clear the net the ball must rise at least 0.33m. if the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racket?

    my answer was found by the equation u^2/2g = d (0.33m)
    so 0.33m x 19.6 ms^2 = 6.47 ms
    the square root of 6.47 ms
    = 2.54 ms

    Is this right?
     
  2. jcsd
  3. Feb 22, 2010 #2
    Hi bd24

    You found the initial vertical velocity, not the initial velocity (which I think the question is asking)
     
  4. Feb 22, 2010 #3
    The ball does have to move at that speed upwards, but the ball left the racket 3 degrees above the horizontal. The speed of the ball was a lot faster than 2.54 ms. How is speed in the Y direction related to the overall speed.
     
  5. Feb 22, 2010 #4
    hmmm, so now i have the speed of the ball in the vertical, is it possible to use trig to solve for the horizontal speed? like 2.54/sin(3°) = 48.5 ms?
    ps. thanks for all the help
     
  6. Feb 22, 2010 #5
    Yes, that's the answer
     
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