The meaning of the ball problem

[Jbreezy]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I don't understand the meaning of this.
$$-16 + \sum_{n=0}^{\infty}(.81)^n= 16+ 32(0.81)\sum_{n=0}^{\infty} 32(.81)^n $$ :editThe one on the left I understand you are just subtracting 16 because the when you start you assume the ball is on the ground. The meaning of the equation on the right is bothering me because I don't understand why you would take n+1 terms and then add 16. I don't understand pictorially what is going on with this one. Why add 16 then take n+1 terms? I don't get it at all. It is bothering me! I don't get it!

 

[Jbreezy]

$$-16 + \sum_{n=0}^{\infty}(.81)^n= 16+ 32(0.81)\sum_{n=0}^{\infty} 32(.81)^n $$

Fixed it I don't know what happened to it above I went to edit it to take an extra 16 out and then it messed the code up. This is what I meant.

 

[Seydlitz]

First it drops 16 feet right? So the distance changes from ##0## to ##16##, then

##d=16+2[16(0.81)+16(0.81)^2+...##

You can see it like this to understand why there's the doubling.

http://www.maplesoft.com/SizedImages/image.ashx?file=38321b83a5fbafd32b9d1470b28c82de_364_280.jpg

Edit:
##n+1## means that you're starting the sum from ##n=1## in the right side, not from ##0##. So it's all good.

 

[Jbreezy]

I feel like I get that (maybe not).
I don't understand this one.

$$16+ 32(0.81)\sum_{n=0}^{\infty} 32(.81)^n$$

Why must you have this one...

$$ 16+$$ 32(0.81)$$\sum_{n=0}^{\infty} 32(.81)^n$$

 

[Seydlitz]

I feel like I get that (maybe not).
I don't understand this one.

$$16+ 32(0.81)\sum_{n=0}^{\infty} 32(.81)^n$$

Why must you have this one...

$$ 16+$$ 32(0.81)$$\sum_{n=0}^{\infty} 32(0.81)^n$$

Because that means $$\sum_{n=0}^{\infty} 32(0.81)^{n+1}$$

Like this
##d=16+2[16(0.81)^1+16(0.81)^2+...##

If I started with n=0 it will look like this:
##d=16+2[16+16(0.81)^1+16(0.81)^2+...##

EDIT: By this I mean if you have started it without (n+1) and just n, and starting from n=0.

I hope you can understand it.

 

[Jbreezy]

$$\sum_{n=0}^{\infty} 32(0.81)^{n+1}$$

I guess I don't get it because it still says that it is n = 0 under the summation symbol.

 

[Seydlitz]

$$\sum_{n=0}^{\infty} 32(0.81)^{n+1}$$

I guess I don't get it because it still says that it is n = 0 under the summation symbol.

Yes ok, you just plug ##0## to ##n+1##, and see what do you get? ##1## right?

so you have ##32(0.81)^{0+1}=32(0.81)^1=32(0.81)^1## for the very first term

*The statement is equivalent to starting the sum from 1 like this:
$$\sum_{n=1}^{\infty} 32(0.81)^{n}$$

We want to have this kind of series:
##d=16+2[16(0.81)^1+16(0.81)^2+...##

 

[Jbreezy]

$$16(0.81)^{0+1}=16(0.81)^1=16$$

What?

If it was ...
$$16(0.81)^0=16$$

I don't see why it would be equal to 16

 

[Seydlitz]

$$16(0.81)^{0+1}=16(0.81)^1=16$$

What?

If it was ...
$$16(0.81)^0=16$$

I don't see why it would be equal to 16

I have edited the previous posts to clear things up, sorry for the mistake.

 

[haruspex]

$$-16 + \sum_{n=0}^{\infty}(.81)^n= 16+ 32(0.81)\sum_{n=0}^{\infty} 32(.81)^n $$

I assume you mean $$-16 + 32\sum_{n=0}^{\infty}(.81)^n= 16+ 32(0.81)\sum_{n=0}^{\infty} (.81)^n $$
Right?
$$-16 + 32\sum_{n=0}^{\infty}(.81)^n= -16+ 32[(0.81)^0+\sum_{n=1}^{\infty} (.81)^n] $$
$$= -16+ 32(0.81)^0+32\sum_{n=1}^{\infty} (.81)^n $$
$$= -16+ 32+32\sum_{n=1}^{\infty} (.81)^n $$
$$= 16+32(.81)\sum_{n=0}^{\infty} (.81)^n $$
Having gone through that, you have $$-16 + S = -16 + 32\sum_{n=0}^{\infty}(.81)^n = 16+(.81)S$$. Solve for S.

 

[Jbreezy]

OK I guess I'm not asking my question right. I want to know what the ball is doing when you use the one on the right (see original post). I can picture what the ball is doing with the other one and not this one. I want to know what the deal is.

 

[haruspex]

OK I guess I'm not asking my question right. I want to know what the ball is doing when you use the one on the right (see original post). I can picture what the ball is doing with the other one and not this one. I want to know what the deal is.

It doesn't have to have a natural physical interpretation to be true. However, it does.
First, let's agree that what you posted was wrong . It should be ##16+ 32(0.81)\sum_{n=0}^{\infty} (.81)^n ##, right? You can read that as "falls 16, then goes through an infinite number of bounces in which the first reaches a height of 16*0.81, and each subsequent is 0.81 times as high as its predecessor."

 

[Jbreezy]

Yeah
I meant

$$-16 + 32\sum_{n=0}^{\infty}(.81)^n= 16+ 32(0.81)\sum_{n=0}^{\infty} (.81)^n$$
Forgot the 32 thanks.

You can read that as "falls 16, then goes through an infinite number of bounces in which the first reaches a height of 16*0.81, and each subsequent is 0.81 times as high as its predecessor."

I'm sorry the whole n+1 thing is really bothering me. I feel like this is a little easier to understand physically.
$$-16 + 32\sum_{n=0}^{\infty}(.81)^n$$
This is just start the ball from the ground and bounce it an infinite number of times. But remember it started from the ground and the question says it was dropped so you subtract 16 because that is the ground to the height of 16 which the ball never took.

I understand what your saying and I guess I understand the the algebra but since there is a physical interpretation of this why do they take n+1 terms. It just makes no sense to me why they would do this? I don't get it. Sorry I just don't get it. The first one sits OK with me.

 

[Seydlitz]

I understand what your saying and I guess I understand the the algebra but since there is a physical interpretation of this why do they take n+1 terms. It just makes no sense to me why they would do this? I don't get it. Sorry I just don't get it. The first one sits OK with me.

We want to know the total distance right? I guess you already understand what happens with the ball.

The total distance is calculated using this series:
$$d=16+2[16(0.81)^1+16(0.81)^2+...$$
That is equal to this:
$$16+\sum_{n=0}^{\infty} 32(0.81)^{n+1}=16+\sum_{n=1}^{\infty} 32(0.81)^{n}$$

It's just a manipulation of the summation symbol. That is the reason the power is (n+1) because you start the summation from 0, whereas you want it to start from 1.