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Homework Help: Linear air resistance

  1. May 20, 2005 #1
    My question deals with times of flight under a linear model of air resistance. Let me describe the problem.

    Up problem: v' = -g - kv, x(0) = 0, v(0) = v_0
    Down problem: v' = -g + kv, x(0) = h, v(0) = 0

    Define T_{up} and h (max. height) by the Up problem: x(T_{up}) = h, v(T_{up}) = 0

    Define T_{down} by Down problem: x(T_{down}) = 0

    Question: Does T_{down} = T_{up}? (prove/disprove)
     
  2. jcsd
  3. May 20, 2005 #2

    arildno

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    Well, what have you done so far?
     
  4. May 20, 2005 #3
    I've solved the problem and would like to see if someone gets the same results.
     
  5. May 20, 2005 #4

    arildno

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    Well, post your solution procedure, and we'll check if it is correct.
     
  6. May 20, 2005 #5
    Let me finish writing it up and then I'll submit it in pdf form.
     
  7. May 20, 2005 #6
    Here are the results for linear air resistance. The quadratic resistance results are being written up now. I hope I did the attachment correctly.
     
  8. May 20, 2005 #7
    I don't think the file uploaded, so here's a copy:

    http://caccmath.cacc.cc.al.us/air-resist.pdf [Broken]
     
    Last edited by a moderator: May 2, 2017
  9. May 20, 2005 #8

    arildno

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    That's some assignment..

    There is, however, one MAJOR flaw in your reasoning, I'll give you the proper solution here:

    1) Air resistance is ALWAYS in the opposite direction of the velocity!!!

    Thus, in both the up&down case, with gravity acting in the negative direction, we have:
    [tex]\frac{dv}{dt}=-g-\frac{k}{m}v (1)[/tex]
    Note that in the "down" case, v<0, that is, air resistance works upwards, as it should.

    Note therefore, that the terminal velocity (i.e, no acceleration) will be: [tex]v=-\frac{mg}{k}[/tex]

    Rearranging (1), and multiplying with the integrating factor [tex]e^{\frac{kt}{m}}[/tex] we have:
    [tex]\frac{d}{dt}(v(t)e^{\frac{kt}{m}})=-ge^{\frac{kt}{m}}[/tex]
    Or, integrating this from t=0 to an arbitrary time, we get:
    [tex]v(t)e^{\frac{kt}{m}}-v(0)=-\frac{mg}{k}e^{\frac{kt}{m}}+g[/tex]
    That is:
    [tex]v(t)=-\frac{mg}{k}(1-e^{-\frac{kt}{m}})+v(0)e^{-\frac{kt}{m}}[/tex]

    Do you follow this?
     
  10. May 20, 2005 #9
    Thanks - the T_{up} = T_{down} was counter to my physical intuition, so I was suspicious. Thanks for your time. This was not an assignment.
     
  11. May 20, 2005 #10

    arildno

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    Welcome to PF!
     
  12. May 20, 2005 #11
    So how did you get the TeX looking formulas in your post?
     
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