# Bouncing ball lab

1. Nov 12, 2007

### ZincPony

1. The problem statement, all variables and given/known data

Figure out the total distance traveled by a ball bouncing vertically when its dropped from a height of 6 ft above the ground and the time the ball was in motion for.

G 32ft/s^2

ball one dropped from 72 inches bounced back up 53 inches with a Coefficient of restitution of .736

2. Relevant equations
the distance i figured would be the CR of each bounce up times 2, with the initial drop the ball traveled 178inches with just one bounce.

72 drop down + (2(53 bounce[up + down])) + (2(53xCR)) + ....

How do i figure out how to calculate the distance without punching in the CR and calculating each bounce up and down

3. The attempt at a solution

I got the Coefficient of restitution for each ball tested, but im having some trouble with figuring out the time and distance problem.

2. Nov 12, 2007

### ZincPony

here are the equations that were presented with the lab.

http://img228.imageshack.us/img228/5433/bouncedm1.jpg [Broken]

Last edited by a moderator: May 3, 2017
3. Nov 12, 2007

### ZincPony

Bouncing ball distance

edit

Last edited: Nov 13, 2007
4. Nov 13, 2007

### rcgldr

Ignoring deformation, the ball bounces an infinite number of times, but in a finite time. What you're looking for is this sum:

$$\lim_{n \rightarrow \infty} \ 72 + 2 \times \ 53 \times \ \sum_{i=0}^n \ (.736)^i$$

This sum can can be calculated with a somewhat clever method.

Last edited: Nov 13, 2007
5. Nov 13, 2007

### catkin

6. Nov 13, 2007

### ZincPony

hmmm soo how would i go about figuring out the clever method to get he sum. im just having a hard time with this, someone please help.

Last edited: Nov 13, 2007
7. Nov 13, 2007

### ZincPony

wouldnt it be like this?? the equation

8. Nov 13, 2007

### catkin

9. Nov 13, 2007

### ZincPony

i looked at the wiki page but im just having some trouble with set up of the equation. i know jeff reid set up an equation for me but i am still having trouble understanding the set up and calculating it.

10. Nov 13, 2007

### Shooting Star

The sum of an infinite geometric series with 1st term 'a' and CR 'r' = a/(1-r), if mod(r)<0. That was what Jeff Reid was talking about. Now try out the formula he gave.

11. Nov 13, 2007

### catkin

Starting with your
S = 72 drop down + (2(53 bounce[up + down])) + (2(53xCR)) + ....
and re-writing it as
S = 72 + 2(72)x0.736 + 2(72)x0.736^2 + 2(72)x0.736^3 + 2(72)x0.736^4 ...
= -72 + 2(72) + 2(72)x0.736 + 2(72)x0.736^2 + 2(72)x0.736^3 + 2(72)x0.736^4 ...
-72 + a + ar + ar^2 + ar^3 ...
where a is 2(72) and r is 0.736

Check out the wikipedia link and you will find that the sum of
a + ar + ar^2 + ar^3 ...
is
a / (1 - r)

12. Nov 13, 2007

### ZincPony

awesome, thanks guys

would i go about the same way to figure out the total time in seconds the ball was in motion.

13. Nov 13, 2007

### catkin

Try it!

What is the time for the first bounce, second bounce, third bounce ... any pattern like ar^n wher n is 1,2,3, ... ?

Last edited: Nov 13, 2007
14. Dec 6, 2007

### ZincPony

wouldnt i just go about finding the time by first finding hte total distance in feet, then figure out the time it was in travel by the feet per second it travels.

total distance / 32ft/s^2
sqrt of (total distance / (32ft/s^2))

15. Dec 7, 2007

### Shooting Star

The ball does not have a constant speed...

16. Dec 11, 2007

### ZincPony

if the balls speed changes thus making it not have equal speed between bounces taking different time how do i go about figuring it out with geometric progression

17. Dec 11, 2007

### Shooting Star

Each time the ball bounces back with 0.736 of the speed with which it hits the ground. We know the time a ball takes to fall back to the ground if it's projected upward with velo v. Sum the times of each bounce-- that's also a GP. Remember, the very first time it only fell.

18. Dec 12, 2007

### catkin

Like I said, "What is the time for the first bounce, second bounce, third bounce ... any pattern like ar^n wher n is 1,2,3, ... ?"