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Help! 2-dimensional motion, with a bounce!

  1. Nov 2, 2004 #1
    If a ball is launched from a launcher at a height h and an angle theta and it lands x meters away from its original x position, and bounces, how can I figure out the angle at which it bounces? How does it relate to what I know? And how do I find the new y velocity?
  2. jcsd
  3. Nov 3, 2004 #2
    You should find the launch angle with the arctan of h/(x/2). If you have no air drag the landing angle is the same. So, if you find the angle you can find the x and y velocities when the ball lands. If there is no friction the x velocity remains the same. To find y velocity you must know if the bounce is perfectly elastic or not. If it is, then the new y velocity is the landing velocity. If not, you should have some bounce efficiency and so you can easiliy find new y velocity. Then you have x and y and you find the angle.
  4. Nov 3, 2004 #3


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    Assuming that the ball lands at the same height at which it started, then the angle at which it hits the pavement is the same as the initial angle! Furthermore, the angle at which it bounces will be the same angle: theta.

    (That's also assuming no air resistance and that the "bounce" is perfectly elastic. If neither of those is true then you can't do the problem without knowing the coefficients of resistance and elasticity.)

    In that case, by conservation of energy, the velocity y when the ball bounces is also exactly the same as the initial velocity!
  5. Nov 3, 2004 #4
    Thank you!!
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