Ball rebounding to less than initial height

AI Thread Summary
A homework problem involves a 0.50 kg ball that rebounds to a height of 20 meters after falling from 30 meters. The ball's impact velocity was calculated to be 24.261 m/s, and it needed an initial velocity of 19.81 m/s to reach the rebound height. The average force exerted on the ball during the 0.002 seconds of contact with the ground was determined to be 11,020 Newtons. The calculations were confirmed by other participants in the discussion. The problem-solving approach and results were well-received, indicating a successful understanding of the physics involved.
MattF
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Hi, I have a homework problem that's giving me a bit of trouble. I hope someone can help me out and verify whether I'm correct or not :)

After falling from rest at a height of 30 meters, a 0.50 kilogram ball rebounds upward, reaching a height of 20 meters. If the contact between ball and ground lasted 0.002 seconds, what average force was exerted on the ball?

Ok, here is how I went about solving this. Using free-fall equation final velocity squared = initial velocity squared + (2 * gravity * height) (g=9.81), I determined that the ball impacted the ground at 24.261 m/s.

Then, using the same equation I determined that to reach a height of 20 meters it had to leave the ground at an initial velocity 19.81 m/s.

Thus, for average force:

F=ma=(mass*change in velocity)/time=[0.5 kg*(19.81 m/s - (-24.261 m/s)]/0.002 seconds

Force= 11020 Newtons

Phew, am I correct?
 
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Welcome to PF!
Looks good to me..
 
arildno said:
Welcome to PF!
Looks good to me..

Wow, I think you have a new medal:


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I've tried to make a Norwegian banner, but it's a bit difficult :biggrin: .
 
Thank you..:shy:
I'll start canvassing for your next one in return..
 
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