- #1
Michael1974
- 4
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1. What forces and impulse impact on a bouncing basketball?
I understand that it is the gravitational force, when it falls, then the normal force when it is on the way back. Are there any additional powers?
The momentum, I came to the potential energy = kinetic energy, where I got the speed. The impulse I got through F = 0.81 * 9.82 and t = 0,34s = 2,70kgm / s. I have observed that the ball bounces about 43% of the height at which it is released from. How can I link to such impulse and power?
But what shall I write in the discussion? Why the result as it gets, aiming for an A, it must have the nuance and detail, but gets nowhere ..2. mv^2=mgh
p = mv
Imp = F*t
F = mg
3. I figured out the momentum via mv^2=mgh , where I got the speed. I got the impulse through F = 0.81 * 9.82 and t = 0,34s = 2,70kgm / s. I have observed that the ball bounces about 43% of the height at which it is released from. I must discuss why the result is as it is? I truly have no idea.
I understand that it is the gravitational force, when it falls, then the normal force when it is on the way back. Are there any additional powers?
The momentum, I came to the potential energy = kinetic energy, where I got the speed. The impulse I got through F = 0.81 * 9.82 and t = 0,34s = 2,70kgm / s. I have observed that the ball bounces about 43% of the height at which it is released from. How can I link to such impulse and power?
But what shall I write in the discussion? Why the result as it gets, aiming for an A, it must have the nuance and detail, but gets nowhere ..2. mv^2=mgh
p = mv
Imp = F*t
F = mg
3. I figured out the momentum via mv^2=mgh , where I got the speed. I got the impulse through F = 0.81 * 9.82 and t = 0,34s = 2,70kgm / s. I have observed that the ball bounces about 43% of the height at which it is released from. I must discuss why the result is as it is? I truly have no idea.
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