What Happens to a Ball's Energy When Thrown Upwards?

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The discussion centers on the physics of a ball thrown upwards, specifically the work done by forces during its ascent. The force exerted by the hand (F) acts over distance d, imparting kinetic energy to the ball, while gravity (mg) does negative work over the total distance (d+s) as the ball ascends and eventually stops. The consensus is that the hand's force only contributes to the ball's energy during the initial push, and once released, the ball's momentum continues until gravity decelerates it. The work-energy theorem is a key concept in understanding these dynamics.

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rudransh verma
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What is going on when the ball is thrown up in the sky. It is pushed by a force F for some distance d. Then the object travels a distance s up in the sky before finally coming to a stop. So what is going on here? Is the force doing work for distance d or distance s (s>d)?
I think change in kinetic energy is due to a force. So even when the momentary push is done the force is still acting all the way to the top and the work done by the force is for distance d. Right?
 
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Person does work ##w_1 = Fd##, weight does work ##w_2 = -mg(d+s)##, and ##w_1 + w_2 = 0## for the ball comes to rest at the highest point.
 
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ergospherical said:
Person does work ##w_1 = Fd##, weight does work ##w_2 = -mg(d+s)##, and ##w_1 + w_2 = 0## for the ball comes to rest at the highest point.
Why not ##w_1=F(d+s)##? Force is also doing work for s distance.
 
rudransh verma said:
Why not ##w_1=F(d+s)##? Force is also doing work for s distance.
Not really, because force of person stops once it leaves their hands.
 
ergospherical said:
Not really, because force of person stops once it leaves their hands.
But pushing the ball up is changing its kinetic energy.
 
Muscles-hand doing work for distance d only.
That means that the ball is gaining energy only during that time in form of impulse.
For distance s, the ball has momentum, which is how that previously gained energy manifests itself.

Please, see:
https://en.m.wikipedia.org/wiki/Impulse_(physics)

:)
 
Lnewqban said:
Muscles-hand doing work for distance d only.
That means that the ball is gaining energy only during that time in form of impulse.
I don’t understand. Why is the s distance situation different from d distance situation?
 
rudransh verma said:
I don’t understand. Why is the s distance situation different from d distance situation?
What is the definition of a force?

https://en.m.wikipedia.org/wiki/Force

:)
 
Lnewqban said:
What is the definition of a force?

https://en.m.wikipedia.org/wiki/Force

:)
Push or pull which causes acceleration.
Force is applied by the hand because we are changing the momentum in some time. Force is also acting on the body when it leaves the hand. So?
 
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  • #10
rudransh verma said:
Push or pull which causes acceleration
Why?
Check the link given above.

Is the ball increasing its velocity after losing contact with the hand?
 
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You could use work-energy theorem to calculate and resolve this.
rudransh verma said:
Push or pull which causes acceleration.
Force is applied by the hand because we are changing the momentum in some time. Force is also acting on the body when it leaves the hand. So?
Clearly the force is no longer acting on the ball when it leaves the hand. Why are you saying it still is?

You know, you could use work-energy to calculate all this, right?
 
  • #12
Lnewqban said:
Is the ball increasing its velocity after losing contact with the hand?
So the ball starts deaccelerating as soon as it leaves the hand since net force is downwards.
Lnewqban said:
Muscles-hand doing work for distance d only.
That means that the ball is gaining energy only during that time in form of impulse.
For distance s, the ball has momentum, which is how that previously gained energy manifests itself.
First we do work for distance d increase its KE and increase its momentum then as soon as it leaves the gravity does opposite work and decrease its kinetic energy and momentum again to zero. If there were no gravity the body would cruise as soon as it would leave the hand with constant momentum and kinetic energy.
 
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  • #13
rudransh verma said:
So the ball starts deaccelerating as soon as it leaves the hand since net force is downwards.

First we do work for distance d increase its KE and increase its momentum then as soon as it leaves the gravity does opposite work and decrease its kinetic energy and momentum again to zero. If there were no gravity the body would cruise as soon as it would leave the hand with constant momentum and kinetic energy.
Excellent! 👍👍👍

animation.gif
:)
 
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  • #14
rudransh verma said:
What is going on when the ball is thrown up in the sky. It is pushed by a force F for some distance d. Then the object travels a distance s up in the sky before finally coming to a stop. So what is going on here? Is the force doing work for distance d or distance s (s>d)?
There are two forces involved here. The force of the throw pushes the ball upward for some distance d. The force of gravity pulls the ball downward over some distance s. This is all very straightforward. What's the problem?
 
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