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Projectile motion - velocity/time unknown

  1. Jan 8, 2009 #1
    A batter hits a pitched ball at a height of 4 feet above the ground so that its angle of projection is 45 degrees and its horizontal range is 350 feet.
    The ball is fair down the left field line where a 24 foot high fence is located 320 feet from home plate.
    By how much will the ball clear the fence?

    vfy=viy+at
    (vf)[tex]^{2}[/tex]= (vi)[tex]^{2}[/tex]+2aD (..we'll say D= displacement)
    D= vi(t)+.5(a)(t[tex]^{2}[/tex])

    Vx=Dx(t)

    Well, I only really know my x displacement, my acceleration constant, and that my initial y velocity and my x velocity are the same because of the 45 degree triangle [at least that's what my teacher said]...

    So, help?! It's been bugging me for days.
     
  2. jcsd
  3. Jan 8, 2009 #2
    Once you have your components of initial velocity along X and Y, X (horizontal) and Y (vertical) equations of motion are only related by time since there is no force along X and the gravitational force along Y, and no constraints on the object like a plane or a string, etc. (it's in free-fall) ... therefore:

    First find the time when the ball is passing the fence.

    Using this time, find the Y location (the vertical displacement).

    Since the fence is only so tall, you can know if it clears and by how much.

    P.S. Welcome to the forums!
     
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