Throwing a baseball up in the air

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The discussion centers on the acceleration of a baseball thrown straight up compared to its acceleration after leaving the hand. Initially, it is suggested that the acceleration is greater while the ball is being thrown due to the rapid increase in velocity. However, it is clarified that if the ball is raised at a constant velocity before being thrown, its acceleration just before launch could be zero. After release, the only force acting on the ball is gravity, which provides a constant downward acceleration. The conclusion emphasizes that the comparison of accelerations depends on the assumptions made about the throwing motion.
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Homework Statement


You throw a baseball straight up in the air so that it rises to a maximum height much greater than your height. Is the magnitude of the acceleration greater while it is being thrown or after it leaves your hand? Explain.


2. The attempt at a solution
I'm guessing that the magnitude of the acceleration is greater while it is being thrown. Acceleration is dv/dt. When I throw a ball straight up, velocity goes from 0 to Vmax in less than a second. When the ball leaves my hand, it takes more than a second for the velocity to reach zero then fall back to the ground. Is this correct?
 
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cdotter said:

Homework Statement


You throw a baseball straight up in the air so that it rises to a maximum height much greater than your height. Is the magnitude of the acceleration greater while it is being thrown or after it leaves your hand? Explain.2. The attempt at a solution
I'm guessing that the magnitude of the acceleration is greater while it is being thrown. Acceleration is dv/dt. When I throw a ball straight up, velocity goes from 0 to Vmax in less than a second. When the ball leaves my hand, it takes more than a second for the velocity to reach zero then fall back to the ground. Is this correct?
Your analysis is quite correct, provided the acceleration is uniform while in contact with your hand. But consider this: If the ball is being raised by your arm at constant upward velocity just prior to being thrown, what is the acceleration of the ball just before launch? After being launched what force acts on the ball? What is the ball's acceleration after launch?

AM
 
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Andrew Mason said:
Your analysis is quite correct, provided the acceleration is uniform while in contact with your hand. But consider this: If the ball is being raised by your arm at constant upward velocity just prior to being thrown, what is the acceleration of the ball just before launch? After being launched what force acts on the ball? What is the ball's acceleration after launch?

AM
So the magnitude of acceleration after the ball releases from my hand (from gravity) is greater than the magnitude of acceleration with the ball in my hand (from being thrown)?
 
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cdotter said:
So the magnitude of acceleration after the ball releases from my hand (from gravity) is greater than the magnitude of acceleration with the ball in my hand (from being thrown)?
It depends on your assumption of how it is thrown.

If one assumes that the throw is made with a constantly accelerating arm, your first answer is correct. But you should state that as your assumption.

I was simply pointing out that if the acceleration is not constant it may not be the case that the magnitude of the acceleration before release is greater than that after. (for example, if the arm accelerates in the first part of the windup but then moves at a fairly uniform speed afterward but before the ball is released).

AM
 
Andrew Mason said:
It depends on your assumption of how it is thrown.

If one assumes that the throw is made with a constantly accelerating arm, your first answer is correct. But you should state that as your assumption.

I was simply pointing out that if the acceleration is not constant it may not be the case that the magnitude of the acceleration before release is greater than that after. (for example, if the arm accelerates in the first part of the windup but then moves at a fairly uniform speed afterward but before the ball is released).

AM

Oh, I understand now. :smile:
 
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