# When the ball is thrown up

Gold Member
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?

• Lnewqban

ergospherical
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.

• russ_watters
Gold Member
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.

ergospherical
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.

Gold Member
Not really, because force of person stops once it leaves their hands.
But pushing the ball up is changing its kinetic energy.

Homework Helper
Gold Member
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.

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

:)

Gold Member
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?

Homework Helper
Gold Member
Gold Member
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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Homework Helper
Gold Member
Push or pull which causes acceleration
Why?

Is the ball increasing its velocity after losing contact with the hand?

Mentor
You could use work-energy theorem to calculate and resolve this.
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?

Gold Member
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.
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.

• ergospherical and Lnewqban
Homework Helper
Gold Member
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! 👍👍👍 :)

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• rudransh verma
Mentor
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?

• jim mcnamara and russ_watters