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Distance to reach terminal velocity

  1. Dec 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A body with a 100 Kg of mass is falling , what's the distance and time to reach terminal velocity?

    2. Relevant equations
    a=dv/dt v=ds/dt drag force= 100v^2
    F=ma

    3. The attempt at a solution
    I don't really need the numbers , just the expressions
    I don't know how to write the equations so I am going to capture my attempt and post as photos. I don't know if I am right or wrong yet. The photos were mixed during the upload , the last photo is the first one and vice versa
     

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  2. jcsd
  3. Dec 30, 2013 #2

    rude man

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    Sorry, x = infinity, t = infinity.

    The terminal velocity is a finite number but the distance and time to reach it are asymptotes which are never completely reached.
     
  4. Dec 30, 2013 #3
    yea i know , but we can get an approximation , i could use excel to do it , but i saw this thread ( https://www.physicsforums.com/showthread.php?t=502933 ) he used differential equations to solve for time , so i thought maybe i could use it to solve for distance
     
    Last edited: Dec 30, 2013
  5. Dec 30, 2013 #4

    rude man

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    You tell me what constitutes "reaching terminal velocity", like for example 99% of terminal velocity, and maybe I can show you how to get that.
     
  6. Dec 30, 2013 #5
    Thanks , i did a spreadsheet and i got the answers , my question is : is this calculus approach wrong?
     
  7. Dec 31, 2013 #6

    haruspex

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    For the time, yes, though my preference would be to keep all the variables as symbols (m for mass, D for drag coefficient etc.) rather then inserting numerical values som early.
    You can solve the time integral using partial fractions.
    For the distance, the easiest way is to use dv/dt = (dv/ds)(ds/dt) = v dv/ds. That gives you a very easy integral.
     
  8. Dec 31, 2013 #7
    So integration does work ??? thanks for that ultra-easy integral :thumbs: , the one i figured out was a nightmare
     
  9. Dec 31, 2013 #8
    if i had a 50 m/s terminal velocity for example , if i put v= 49 m/s , i can work it out with calculus right?
     
  10. Dec 31, 2013 #9

    rude man

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    Right!
     
  11. Dec 31, 2013 #10
    Thanks
     
  12. Dec 31, 2013 #11
    One last question before we call it a day , when i use differential equation calculator in wolfram , he used log instead of ln in the integration , he gave me t in terms of v . when he was trying to put v the subject ( to put v in terms of t ) he used e . How come?
     
  13. Dec 31, 2013 #12

    rude man

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    What was your diff. equation?
     
  14. Dec 31, 2013 #13
    Int(t)=int(dv/(228-0.113v^2))
     
  15. Dec 31, 2013 #14

    rude man

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    OK, you mean dt = dv/(228 - 0.113 v2).

    That "228" bothers me. I have g for that constant = 9.81 ms-2.
    Also, you need to know the drag coefficient = drag force/v2. I saw nothing in your problem that gave that coefficient a number. Where did you get the number 0.113 from, which has to be the drag coefficient divided by the mass of the diver?
     
  16. Dec 31, 2013 #15
    This is not really about skydiving, i am pretty sure the other calculation are correct, the body is teardrop shape which according to google has a coefficient of drag of 0.05
     
  17. Dec 31, 2013 #16

    rude man

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    OK, but I still think you should show how you got those two numbers: 228 and 0.113.

    If you're sure they're right (and I don't think they are), then go ahead and show what your solution was for v(t) or t(v) or both.
     
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