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The meaning of the ball problem

  1. Jul 23, 2013 #1
    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



    3. The attempt at a solution

    I don't understand the meaning of this.



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


    The 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!
     
    Last edited by a moderator: Jul 24, 2013
  2. jcsd
  3. Jul 23, 2013 #2
    $$-16 + \sum_{n=0}^{\infty}(.81)^n= 16+ 32(0.81)\sum_{n=0}^{\infty} 32(.81)^n $$

    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.
     
  4. Jul 23, 2013 #3
    Last edited: Jul 23, 2013
  5. Jul 23, 2013 #4
    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)$$\sum_{n=0}^{\infty} 32(.81)^n$$
     
  6. Jul 23, 2013 #5
    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.
     
    Last edited: Jul 23, 2013
  7. Jul 23, 2013 #6
    $$\sum_{n=0}^{\infty} 32(0.81)^{n+1}$$

    I guess I don't get it because it still says that it is n = 0 under the summation symbol.
     
  8. Jul 23, 2013 #7
    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+...##
     
    Last edited: Jul 23, 2013
  9. Jul 23, 2013 #8
    $$16(0.81)^{0+1}=16(0.81)^1=16$$

    What?

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

    I don't see why it would be equal to 16
     
  10. Jul 23, 2013 #9
    I have edited the previous posts to clear things up, sorry for the mistake.
     
  11. Jul 24, 2013 #10

    haruspex

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    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.
     
  12. Jul 24, 2013 #11
    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.
     
  13. Jul 24, 2013 #12

    haruspex

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    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."
     
  14. Jul 24, 2013 #13
    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.


    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.
     
  15. Jul 24, 2013 #14
    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.
     
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