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Hey guys, I need help with graphing acceleration to time for a bouncing ball (1 Viewer)

bleakarcher

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Hey guys, can someone draw and explain this sort of graph for me?
 

IamBread

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The acceleration? It's going to be g. not going to change.
 

RealiseNothing

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I'm not doing moving about(I assume bleak is), but when the ball bounces, wouldn't it have to go down to a velocity of 0 then accelerate again as it bounces back up? And the bounce back up isn't 9.81ms^-1 is it?

Just curious.
 

deswa1

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You are right that velocity changes as a function of time but acceleration is always constant. Regardless of where the ball is, the gravitational force downwards will lead to a gravitational acceleration downwards of 9.8ms^-1, thus acceleration is constant.
 

RealiseNothing

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You are right that velocity changes as a function of time but acceleration is always constant. Regardless of where the ball is, the gravitational force downwards will lead to a gravitational acceleration downwards of 9.8ms^-1, thus acceleration is constant.
Ah I see, thanks for clearing it up.
 

bleakarcher

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Apparently, according to my teacher, as the ball strikes the ground and is sent back up again in that extremely small interval of time the acceleration is a large positive value if you taking gravity acting negatively. WTF?
 

Timske

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The acceleration of the ball while it is in the air before it hits the ground is -g (downward) so when it bounces back up its going opposite direction to -g so its +g
 

bleakarcher

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The acceleration of the ball while it is in the air before it hits the ground is -g (downward) so when it bounces back up its going opposite direction to -g so its +g
No it should still be -g as it is going upwards, thats why it comes back down lol.
 

Timske

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oops.. velocity is positive going downwards and once it hits the ground velocity = 0 and backup - velocity. At the moment before it hits the ground
the balls has gained momentum during its time in the air so all this would be reflected upwards
 

Nooblet94

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The moment the ball hits the ground there's an acceleration in the upward direction, larger than the acceleration due to gravity. Apart from that it's constant.

EDIT: Just to clarify, that upward acceleration is instantaneous.
 

Carrotsticks

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The acceleration? It's going to be g. not going to change.
-g.

You are right that velocity changes as a function of time but acceleration is always constant. Regardless of where the ball is, the gravitational force downwards will lead to a gravitational acceleration downwards of 9.8ms^-1, thus acceleration is constant.
*ms^-2.

ms^-1 is velocity.

Apparently, according to my teacher, as the ball strikes the ground and is sent back up again in that extremely small interval of time the acceleration is a large positive value if you taking gravity acting negatively. WTF?
The forces acting upon the ball at all times is -g ms^-2 due to gravity.

However upon collision with the ground, a phenomenon called 'Elastic Collision' occurs, causing the momentary upward acceleration. I believe this is what your teacher was pertaining to.

http://en.wikipedia.org/wiki/Elastic_collision
 

Carrotsticks

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Depends which way you set as positive :p
Indeed, though I think it would make sense to make the ground negative and the sky positive.

I don't understand all this riff raff about changing the positive direction unless it significantly makes computations easier (like in Extension 2 Resisted motion problems).
 

IamBread

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Indeed, though I think it would make sense to make the ground negative and the sky positive.

I don't understand all this riff raff about changing the positive direction unless it significantly makes computations easier (like in Extension 2 Resisted motion problems).
I generally do -g, unless it's only going in 1 direction, so if it's a falling ball I'd use g (assuming the ball doesn't bounce up).
 

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