# Bouncing ball series

1. Jul 4, 2013

### Jbreezy

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

A ball is dropped from a height of 16 feet. Each time it drops h feet, it rebounds 0.81h feet. Find the total distance traveled by the ball.

2. Relevant equations

Geometric series sum of ar^n

3. The attempt at a solution

D1= 16 feet
D2 = 16(0.81) + 16(0.81) = 32(0.81)
Then D3 = 32(0.81)^2

So then you have D = 16 + 32(sum from n = 1 to ∞) (0.81)^n+1

= 16 + 32(0.81)(sum from n = 0 to ∞) (0.81)^n
When I first did my own I got D = 16 + 32(sum from n = 0 to ∞)(0.81)^n

My question is why is this not correct?

Also the book did it a different way and had
-16 + (sum from n = 0 to ∞)32(0.81)^n

why would you subtract 16? This is what I had when it was wrong except they subtract 16. Why?

2. Jul 4, 2013

### Nugso

Hello. Sorry but I didn't quite understand. ( That's probably because of my English)

The way your book answers it is correct. Because by summing 32(0,81)^n where n is from 0 to ∞, it assumes that the ball travels D1 twice. ( Up and down)

Here is how it looks like without subtracting 16;

http://www.sketchtoy.com/42921712

http://www.sketchtoy.com/42921326

We have D= 16 + 32(sum from n=1 to ∞)(0,81)n

Last edited: Jul 4, 2013
3. Jul 4, 2013

### Infrared

Because the ball only has one trip of length 16. If the ball started from the ground and bounced up 16 feet, then the total distance would be $2\sum_{n=0}^{\infty} 16(.81)^n= \sum_{n=0}^{\infty} 32(.81)^n$. The answer is 16 less than this since the first trip of length 16 is not made.

4. Jul 4, 2013

### Jbreezy

Oh I see thanks