Acceleration at the time of throwing a ball

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SUMMARY

The discussion centers on the acceleration of a ball thrown vertically, comparing the acceleration during the throw to the acceleration due to gravity (g) after the ball is released. Participants concluded that the acceleration while throwing the ball (a) is greater than g, as the thrower applies a force over a shorter distance, resulting in a higher initial velocity. The work-energy principle is also highlighted, indicating that the work done during the throw must equal the work done against gravity as the ball rises. Thus, the net acceleration during the throw exceeds gravitational acceleration.

PREREQUISITES
  • Understanding of 1D kinematics
  • Familiarity with Newton's laws of motion
  • Basic knowledge of work-energy principles
  • Concept of gravitational acceleration (g)
NEXT STEPS
  • Study the work-energy theorem in classical mechanics
  • Learn about the relationship between force, mass, and acceleration (Newton's second law)
  • Explore kinematic equations for uniformly accelerated motion
  • Investigate the effects of air resistance on projectile motion
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Adjoint
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Homework Statement



(It's just a simple question.)
A ball is thrown straight up in the air so that it rises to a maximum height much greater than the thrower's height. Is the magnitude of the acceleration greater while it is being thrown or after it is thrown?

Homework Equations



1D kinematics.

The Attempt at a Solution



Magnitude of the acceleration after the ball is thrown is g. But I am having trouble in conceptualizing what is happening when the ball is being thrown. Surely when I am throwing the ball I am giving it acceleration. But how can I tell if it is greater or smaller than g?
 
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Do you have a maximum height?
 
No. They just says much higher than the thrower.
 
Or any other information? an interval over which the velocity changed from 0 to the velocity it leaves the hand.
 
Well this is really just a conceptual question. If I exert a force of magnitude mg upwards on the ball, what is the magnitude of the net acceleration of the ball? keep in mind that gravity is still exerting a force on the ball.
 
Adjoint said:
No. They just says much higher than the thrower.

Try using energy. The kinetic energy put into the ball during the throwing process must be the same as the kinetic energy that is removed from the ball as it rises against gravity, right?
 
jbriggs444 said:
The kinetic energy put into the ball during the throwing process must be the same as the kinetic energy that is removed from the ball as it rises against gravity, right?

Yes. The kinetic energy reduces. So velocity after the ball is thrown is less than the velocity with which the ball was thrown. But how do I get to acceleration?
 
I think the question is actually asking if a > g or a < g (where a is the acceleration with which a ball is thrown).
Now how can I tell that if they give no other info?
 
If you have two masses being accelerated to the same speed v but with different accelerations a1 < a2, which mass travels the furthest before reaching the final velocity?
 
  • #10
Adjoint said:
I think the question is actually asking if a > g or a < g (where a is the acceleration with which a ball is thrown).
Now how can I tell that if they give no other info?

You accelerate the ball with acceleration a over a distance comparable with your height.
Due to this acceleration, the ball will have a speed vo.
Then you release the ball and the speed decreases with acceleration g, over a distance much larger than your speed, until the speed becomes zero.

So you have speed going from 0 to vo with a, over d1; and from vo to 0 with g, over d2.
d2>>d1. What can you say about a compared with g?
 
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  • #11
Orodruin said:
If you have two masses being accelerated to the same speed v but with different accelerations a1 < a2, which mass travels the furthest before reaching the final velocity?

As v = at I think I can say if a1 < a2 then to attain same speed v, mass 1 will take longer time.
And by y = 0.5at2, mass 1 will travel further. Right? But maybe not. Because we have a1 < a2.
I am confused.
 
  • #12
Note that the time to reach v is also different for the two accelerations.
 
  • #13
nasu said:
You accelerate the ball with acceleration a over a distance comparable with your height.
Due to this acceleration, the ball will have a speed vo.
Then you release the ball and the speed decreases with acceleration g, over a distance much larger than your speed, until the speed becomes zero.

So you have speed going from 0 to vo with a, over d1; and from vo to 0 with g, over d2.
d2>>d1. What can you say about a compared with g?

Well Now I understand.
Certainly a > g because a needs a shorter distance than g to cause same change in velocity.
 
  • #14
nasu said:
You accelerate the ball with acceleration a over a distance comparable with your height.


I have to disagree, I can throw a baseball straight up, starting about waist high, and releasing it about arm length above my head, so a distance about 1 meter.

If I throw it like that, it'll probably hit about 15-20 meter's high. I wouldn't say that my distance that I accelerate the ball over is comparable with the height it reaches. (And that's with only little league baseball skills ha!)
 
  • #15
There is also an energy argument: the work done by the accelerating forces on the mass has to equal the the negative work done by gravity.

Work done by acceleration: maL, where L is the length of the accelation
Work done by gravity after acceleration: -mgh

Thus gh = aL and since h >> L, a >> g.
 
  • #16
BiGyElLoWhAt said:
I have to disagree, I can throw a baseball straight up, starting about waist high, and releasing it about arm length above my head, so a distance about 1 meter.

If I throw it like that, it'll probably hit about 15-20 meter's high. I wouldn't say that my distance that I accelerate the ball over is comparable with the height it reaches. (And that's with only little league baseball skills ha!)

1 m is comparable with your height. Or average human height.
You can accelerate it over a much shorter distance but you can hardly accelerate it over a distance much larger than your height.

I never said that that the distance you accelerate the ball is comparable with the heigth it reaches. You probably misread my post.
 
  • #17
ahhh that I did, sorry nasu.
 
  • #18
No worries. :)
I am glad I was able to make clear what I meant.
 

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