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Projectile Motion

  1. Oct 10, 2004 #1
    I just started doing some problems and this basic pm problem is giving me a lot of trouble.

    If a ball was throwen horizontally with 45.08 m/s, how far would the ball fall vertically by the time it reached home plate, 18.28m away?

    The initial velocity is 45.08 m/s. Now that is the x and y velocity combined, correct? How do I go about finding the x and y velocity? If I don't need to do it for this particular problem could you explain how to do it anyway and how to solve the problem. I tried using arc-cos to find the initial angle then find the velocities from there but I don't think that's right.

    Thanks for reading. :smile:
  2. jcsd
  3. Oct 10, 2004 #2
    If it's thrown horizontally, the vertical component of its initial velocity is 0. Since there aren't any forces acting on the ball horizontally (ignoring air resistance), it travels at a constant velocity. You can then work out the time taken to travel 18.28 m (t = x/v). What you want to do is find out the distance the ball falls in this time. Use the equation s = ut + (1/2)at^2 (using a = g, u = 0 as explained above and t as the time you've calculated).
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