bleakarcher
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- 2013
Hey guys, can someone draw and explain this sort of graph for me?
Ah I see, thanks for clearing it up.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.
No it should still be -g as it is going upwards, thats why it comes back down lol.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
When a ball hits the ground, the ball gets squished giving it an upward force and so the acceleration isn't exactly instantaneous - just a small amount of timeJust to clarify, that upward acceleration is instantaneous.
-g.The acceleration? It's going to be g. not going to change.
*ms^-2.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.
The forces acting upon the ball at all times is -g ms^-2 due to gravity.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?
Depends which way you set as positive
Indeed, though I think it would make sense to make the ground negative and the sky positive.Depends which way you set as positive![]()
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).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).