Ball rebounding to less than initial height

Click For Summary

Discussion Overview

The discussion revolves around a homework problem involving the physics of a ball rebounding to a height less than its initial drop height. The focus is on calculating the average force exerted on the ball during its contact with the ground after falling from a height of 30 meters and rebounding to 20 meters.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents a problem involving a ball falling from 30 meters and rebounding to 20 meters, seeking verification of their calculations regarding average force.
  • The participant calculates the impact velocity using the free-fall equation and finds it to be 24.261 m/s.
  • They also calculate the initial velocity needed to reach the rebound height of 20 meters as 19.81 m/s.
  • The average force is computed using the change in velocity and the time of contact, resulting in a force of 11020 Newtons.
  • Two other participants respond affirmatively, indicating that the calculations appear correct.

Areas of Agreement / Disagreement

There is general agreement among participants regarding the correctness of the calculations presented, although no detailed critique or alternative viewpoints are offered.

Contextual Notes

Assumptions regarding air resistance and other potential factors affecting the ball's motion are not discussed. The calculations depend on the accuracy of the free-fall equations used and the values for gravity.

Who May Find This Useful

Students working on physics problems related to motion, energy conservation, and force calculations may find this discussion relevant.

MattF
Messages
13
Reaction score
0
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?
 
Physics news on Phys.org
Welcome to PF!
Looks good to me..
 
arildno said:
Welcome to PF!
Looks good to me..

Wow, I think you have a new medal:


! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !
====CONGRATULATIONS===
! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !

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..
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 13 ·
Replies
13
Views
5K
Replies
12
Views
1K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
10K