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Conservation of motion and basketball shot

  1. Oct 6, 2013 #1
    A basketball of mass 0.624 kg is shot from a vertical height of 1.55 m and at a speed of 12.1 m/s. After reaching its maximum height, the ball moves into the hoop on its downward path, at 3.05 m above the ground. Using the principle of energy conservation, determine how fast the ball is moving just before it enters the hoop.

    v=√(v0^2+2g(y0-y))

    I've tried plugging the numbers in an I come up with an answer of 13.26 m/s. However this is incorrect. I believe I might be thinking of where the ball is being shot from differently.

    v=√((12.1^2)+2(9.81)(1.55-3.05)

    Any help would be appreciated. Thanks, Dominic.
     
  2. jcsd
  3. Oct 6, 2013 #2

    phyzguy

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    It looks like you have the formula correct, and you have put the numbers in correctly, but you must have made a mistake doing the calculation. You are taking the 12.1^2, adding a negative number, then taking the square root, so the answer needs to come out less than 12.1 m/s. Check the calculation again.
     
  4. Oct 6, 2013 #3
    Got thanks! I actually just calculated the balls maximum height and then treated it as a free fall problem.
     
  5. Oct 6, 2013 #4

    phyzguy

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    Glad to help. Don't forget the Thanks button!
     
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