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Two dimensional kinematics

  1. Oct 17, 2009 #1
    An athlete throws a basketball upward from the ground, giving it speed 9.7 m/s at an angle of 64.0° above the horizontal.

    (a) What is the acceleration of the basketball at the highest point in its trajectory?
    m/s2

    (b) On its way down, the basketball hits the rim of the basket, 3.05 m above the floor. It bounces straight up with one-half the speed with which it hit the rim. What height above the floor does the basketball reach on this bounce?

    (a) is -9.8m/s2.

    For (b) I found the time when the ball hits the rim 1.3s. Then I found the final velocity -3.04m/s and divided by 2 to get -1.52m/s which is the initial velocity of the ball when it bounces straight up vertically from the rim.

    I am lost at where to go now and need help with solving (b).
     
  2. jcsd
  3. Oct 18, 2009 #2

    rl.bhat

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    Homework Helper

    Your time is correct.
    Check the calculation of the vertical component of the velocity at that time.
     
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