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Ball Problem

  1. Aug 22, 2008 #1

    I have a problem that needs explanation.

    A ball is thrown straight up in the air, and then falls back down into your hand. What is the ball's acceleration at the top of the throw? i.e when it has just started to come back down again from being thrown up. I assume the acceleration is zero as i think there is no energy at that point...but i think i'm wrong..

  2. jcsd
  3. Aug 22, 2008 #2

    Doc Al

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    Staff: Mentor

    At the top of the throw the velocity is zero, not the acceleration.
  4. Aug 22, 2008 #3
    can you explain why the acceleration isn't zero as well? i think that when the ball is thrown from the hand it goes a positive acceleration as it goes up and a negative acceleration as it goes down, but that point where it goes from going up to going down should the acceleration not also be zero at that point?
  5. Aug 22, 2008 #4


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    If at the top, both velocity and acceleration are 0, this means that the ball would suspended in mid-air, because it's velocity (0) does not change.
  6. Aug 22, 2008 #5
    is it not suspended in mid air for a split second when it reaches the very top of its ascent and begins it descent?
  7. Aug 23, 2008 #6


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    If it's suspended for a split second it means that it's velocity is zero, but not that it's acceleration is zero. If both are zero then it'll be suspended forever (unless it's also possible that the time derivative of acceleration is non-zero at that point).
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