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