# When does the ball come to rest?

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1. Nov 30, 2016

### Gopal Mailpalli

1. The problem statement, all variables and given/known data
A ball is dropped vertically from a height H on to a plane surface and permitted to bounce repeatedly along a vertical line. After every bounce, its kinetic energy becomes a quarter of its kinetic energy before the bounce. The ball will come to rest after time?

2. Relevant equations
Kinetic Energy = $\frac{1}{2}mv^{2}$
Velocity = $\sqrt{2gRh}$ , where R is the fraction of hight ball reaches after every bounce
3. The attempt at a solution
I could recognize the geometrical series and attempted the solution which gave me a wrong time.

2. Nov 30, 2016

### BvU

Did you ? And how did you interpret the sum of this series to get the answer ?

3. Nov 30, 2016

### Gopal Mailpalli

Excuse me! Some where I did a mistake, i assumed the the initial Kinetic energy be K and next consecutive terms are (1/4)K, (1/4)^2 K ...
This represents a Geometric Progression, this is how I approached the problem.

4. Nov 30, 2016

### BvU

Yes, that's what it says in the problem statement. So what series did you get and how did you work the sum of that series around to the answer you gave ?

There is a simple mistake that's easy to make here, but if I ask specifically I give away the answer, which is a bit against PF culture.

5. Nov 30, 2016

### Gopal Mailpalli

I found the coefficient of restitution which is 0.5, found out the height using the respective COR - Height relation and formed a geometric progression. I calculated the total height and went further to solve the time it takes using basic laws of motion.

6. Nov 30, 2016

### PeroK

You have to post your working, equations and all. We can't spot a mistake from a description of what you did.

7. Nov 30, 2016

### BvU

I don't understand what you did. Can you post your working ?
Why bother to find a total height ? The exercise asks for a time.
What total height did you find ? Something like $4h\over 3$ ?

8. Nov 30, 2016

### Gopal Mailpalli

I mean sum of the geometric progression, which i got 4h/3.

9. Nov 30, 2016

### BvU

Nice, but useless: the total time has nothing to do with the time needed to fall 4h/3 (why not ?).

10. Dec 1, 2016

### Gopal Mailpalli

Then how do I approach the problem? I don't have an idea.

11. Dec 1, 2016

### BvU

You will also get a geometric series if you calculate the sum of the times the ball needs to get back to the ground after each bounce.

12. Dec 8, 2016

### Gopal Mailpalli

Still, i couldn't solve the problem.

13. Dec 8, 2016

### haruspex

It's a bit easier to see what is going on if you look at the time between bounces.
Between first and second bounce, how long is it in the air?
What about between second and third bounces?
What is the pattern?