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Impulse from basket ball against ground at an angle

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data

    To basketball player bounces the 0.60-kg basketball to another player by bouncing it off the court. The ball hits the court with a speed of 6.5 m/s at an angle of 50° from the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the court?


    2. Relevant equations

    I know that Impulse is the change in momentum, or the force times the change in time.

    3. The attempt at a solution

    I cannot figure out how to find the Impulse with simply the velocity an angles. Is there a change in momentum if all it does is change direction? If so, how do I use the angles to calculate the change?
    I try doing the x/y components of the velocity but it seems to cancel out.
     
  2. jcsd
  3. Nov 7, 2009 #2

    Delphi51

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    Homework Helper

    You must use xy components. There is no change in the x velocity, so no impulse. But there is a change in the y velocity - it reverses direction so you'll have something like Δv = -v - v = -2v. Of course you know the number for v, the vertical component of the 6.5 m/s.
     
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