Bouncing Ball Lab: Calculate Distance & Time

AI Thread Summary
The discussion revolves around calculating the total distance traveled and time taken by a ball dropped from a height of 6 feet, considering its bounces. The ball's initial drop is 72 inches, and it bounces back up to 53 inches with a coefficient of restitution (CR) of 0.736. Participants explore the use of a geometric series to sum the distances of infinite bounces, leading to the formula for total distance. There is also a focus on determining the time of each bounce, noting that the ball's speed changes with each bounce, complicating the time calculation. The conversation emphasizes the need for a systematic approach to solve these physics problems using the principles of geometric progression.
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Homework Statement



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

Homework 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


The Attempt at a Solution



I got the Coefficient of restitution for each ball tested, but I am having some trouble with figuring out the time and distance problem.
 
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here are the equations that were presented with the lab.

http://img228.imageshack.us/img228/5433/bouncedm1.jpg
 
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Bouncing ball distance

edit
 
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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.
 
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hmmm soo how would i go about figuring out the clever method to get he sum. I am just having a hard time with this, someone please help.
 
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Jeff Reid said:
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.

wouldnt it be like this?? the equation
 
i looked at the wiki page but I am 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
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
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
awesome, thanks guys

would i go about the same way to figure out the total time in seconds the ball was in motion.
 
  • #13
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, ... ?
 
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  • #14
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
The ball does not have a constant speed...
 
  • #16
Shooting star said:
The ball does not have a constant speed...

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
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
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, ... ?"
 
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