# Acceleration at the time of throwing a ball

1. Jun 20, 2014

1. The problem statement, all variables and given/known data

(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?

2. Relevant equations

1D kinematics.

3. 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?

2. Jun 20, 2014

### BiGyElLoWhAt

Do you have a maximum height?

3. Jun 20, 2014

No. They just says much higher than the thrower.

4. Jun 20, 2014

### BiGyElLoWhAt

Or any other information? an interval over which the velocity changed from 0 to the velocity it leaves the hand.

5. Jun 20, 2014

### BiGyElLoWhAt

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.

6. Jun 20, 2014

### jbriggs444

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?

7. Jun 20, 2014

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?

8. Jun 20, 2014

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?

9. Jun 20, 2014

### Orodruin

Staff Emeritus
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. Jun 20, 2014

### nasu

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?

11. Jun 20, 2014

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. Jun 20, 2014

### Orodruin

Staff Emeritus
Note that the time to reach v is also different for the two accelerations.

13. Jun 20, 2014

Well Now I understand.
Certainly a > g because a needs a shorter distance than g to cause same change in velocity.

14. Jun 20, 2014

### BiGyElLoWhAt

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. Jun 20, 2014

### Orodruin

Staff Emeritus
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. Jun 20, 2014

### nasu

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. Jun 20, 2014

### BiGyElLoWhAt

ahhh that I did, sorry nasu.

18. Jun 20, 2014

### nasu

No worries. :)
I am glad I was able to make clear what I meant.