Understanding the Acceleration of a Thrown Ball

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The discussion centers on the acceleration of a ball thrown straight up as it reaches the top of its trajectory. At the peak, while the ball's velocity is indeed zero, its acceleration is not; it remains constant due to gravity, which acts downward at approximately 9.81 m/s². The misconception arises from confusing velocity with acceleration; the ball does not suspend in mid-air, as it continues to experience gravitational pull. Thus, even at the highest point, the ball is not in a state of zero acceleration. Understanding this distinction clarifies that the ball cannot remain suspended indefinitely without an opposing force.
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Hi,

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..

thoughts?
 
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At the top of the throw the velocity is zero, not the acceleration.
 
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?
 
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.
 
is it not suspended in mid air for a split second when it reaches the very top of its ascent and begins it descent?
 
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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