Solving Dropped Tennis Ball Problem: Calculating Force of Floor

  • Thread starter Thread starter J.T.
  • Start date Start date
  • Tags Tags
    Ball Tennis
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a tennis ball dropped from a height and its interaction with the floor upon impact. The original poster is trying to calculate the force exerted by the floor on the ball after it compresses upon impact.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between initial and final velocities, displacement, and the need to find acceleration. Questions arise about the appropriate equations to use for these calculations.

Discussion Status

Some participants have provided guidance on how to approach the problem, particularly in determining acceleration and using kinematic equations. The original poster has made progress but continues to encounter difficulties with subsequent calculations.

Contextual Notes

There is mention of specific values and constraints, such as the maximum compression of the ball and the need for precision in calculations, which may affect the results. The original poster expresses uncertainty about the accuracy of their answers based on feedback from an online program.

J.T.
Messages
8
Reaction score
0
Hello! I was hoping I could receive some help with a certain problem.

A 0.7kg tennis ball is dropped from rest at a height of 5m onto a hard floor.

--From that, I was able to determine that the final velocity is 9.9m/s.

The part I'm having trouble with is this: A flash photograph shows that the ball is compressed a maximum of 0.6cm when it strikes the floor. Assuming that the acceleration of the ball is constant during its contact with the floor, what force does the floor exert on the ball?

I really don't know what to do with this information...
Any help would be greatly appreciated!
 
Physics news on Phys.org
Originally posted by J.T.
The part I'm having trouble with is this: A flash photograph shows that the ball is compressed a maximum of 0.6cm when it strikes the floor. Assuming that the acceleration of the ball is constant during its contact with the floor, what force does the floor exert on the ball?

It means that the ball was brought to a stop in a distance of 0.6cm. Use your final velocity of the first part as the initial velocity of the second part (the final velocity of the second part is obviously zero).

You know:

*Initial Velocity
*Final Velocity
*Change In Displacement

You need:

*Acceleration

There is precicely one equation that relates all of those. Can you find it?
 
Thank you! I was finally able to solve it...
 
...And yet I run into another problem.

After I find the force it asks me, "what time does the force act in bringing the ball to rest?"

I realize I'm supposed to use one of the kinematic equations, so I've been plugging in numbers for a and x, etc. The thing is, all of my solutions are wrong... I'm guessing that I've been plugging in values that aren't supposed to be plugged in...?

Help please? Many thanks in advance!
 
You know the force now so you know the acceleration.

You know that the speed at time t is given by v(t)= v0+ at
(a is positive (up) here, v0= -9.9 m/s) and you are looking for the time until v(t)= 0: that is t= -v0/a.)
 
I keep getting the wrong answer...Grr. :frown:

Earlier, I determined that a=8167.5m/s^2 and f=5717.25N.
The online program I'm using told me that my answer was correct (for force), so I proceeded to solve for time.

No matter how many times I plug the numbers into the equation, I keep getting 0.0012s, which is wrong, according to this program.

Did I make a miscalculation somewhere?
 
My bad.

The program refused to take 0.0012s, but it took 0.00121s.
I guess I just needed to be a little more accurate...(--;)
I'm such an idiot.

Anyway, thank you for all the help!
Now I have an exam to look forward to on Friday :)
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K