Can a Tennis Ball Demonstrate Physics Principles?

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Homework Help Overview

The discussion revolves around a physics problem involving the drop and rebound of a tennis ball. The original poster attempts to calculate the average acceleration of the ball during its contact with the floor after being dropped from a height of 4.93 m and rebounding to a height of 2.68 m.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the ball's velocity before and after the bounce, questioning the use of height values in the equations. There is an exploration of the initial and final positions of the ball during its upward motion after the bounce.

Discussion Status

Participants are actively engaged in checking calculations and clarifying the setup of the problem. There is a recognition of potential errors in the conversion of time and the interpretation of height values, but no consensus has been reached on the correct answer yet.

Contextual Notes

There are indications of confusion regarding sign conventions and the correct application of physics equations, as well as a focus on ensuring accurate unit conversions.

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[SOLVED] Another use for a tennis ball

Homework Statement


To test the quality of a tennis ball, you drop it onto the floor from a height of 4.93 m. It rebounds to a height of 2.68 m. If the ball is in contact with the floor for 12.3 ms, what is its average acceleration during that contact?


Homework Equations





The Attempt at a Solution



v1 = -sqrt(-2(g)(4.93m)) = -9.83 m/s
v2 = sqrt(2(g)(-2.68+4.93)) = 6.64 m/s

Average acel = v2-v1/delta time = (6.64-(-9.83))/12.3x10^(-3)ms = 1.34 m/s^2?

The answer isn't correct but I have no idea where I'm going wrong.
 
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Your value for v2 is wrong. I think the problem is your value for the height in the v2 equation? Why are you using (-2.68+4.93) as the height in that equation?
 
From sqrt(2g(yFinal-yInitial)
 
Think of the ball traveling back up after the bounce as a completely separate part of the problem from the trip downward before the bounce, with it's own initial and final positions.

Now, thinking like this, what is the balls initial position after the bounce? What is its final position? Does this change your answer?
 
0-(-2.68)?
 
Yes, that will give you the correct change in position for the ball after the bounce, though your sign choices are confusing me a tiny bit.

The ball will start at 0 after the bounce, and end at +2.68, thus, yf-yi=2.68.

This is what you got as well, so it's fine. I am just wondering why you used the signs you did.
 
I still end up getting an incorrect answer :( So frustrating!
 
What do you get for v2 this time around? ( Please show the calculation.)
 
sqrt(2*(9.8)(2.68)) = 7.247620299
 
  • #10
Ok. This is what I get. Now what do you get for the avg acceleration, calculation included?
 
  • #11
((7.247620299+9.829954222)/(.00123m/s^2)) = 13884.20693 m/s^2. But its wrong Q_Q
 
  • #12
I get 13878 (rounded of course), not 13884. I used those numbers, though. Check your calculations again.
 
  • #13
Rechecked get the exact same answer.
 
  • #14
Have you converted the time correctly?
 
  • #15
mda said:
Have you converted the time correctly?

Ahh, yes. I see it now too. Check your time conversion. I think you have a decimal place error. Nice catch mda.
 
  • #16
Always something simple! Raaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa. Thanks for the help! Really appreciate it.
 
  • #17
Anytime!
 

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