Can a Tennis Ball Demonstrate Physics Principles?

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The discussion focuses on calculating the average acceleration of a tennis ball during its contact with the floor after being dropped from a height of 4.93 meters. Initial calculations for the rebound height and velocities were incorrect due to confusion over the height values used in the equations. Participants emphasized the importance of treating the downward and upward motions as separate phases, leading to the correct change in position after the bounce. A critical error was identified in the time conversion, which affected the final acceleration calculation. The conversation highlights the collaborative effort to troubleshoot and correct the physics problem, demonstrating the application of physics principles through practical examples.
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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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