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Helpless Free falling body

  1. Sep 4, 2003 #1
    I need an answer for a practical question. Say an unconscious (but rigid) person fell out of a helicopter that was not moving. If you DON'T ignore air resistance and terminal velocity, what would be his displacement after a given time (assuming the initial position is at zero, for simplification). In other words, what is the total distance he would fall after 6 sec, 12 sec, 18 sec, 24 sec, 30 sec, 36 sec, 42 sec, 48 sec, 54 sec, 60 sec, and so on?
    Last edited: Sep 4, 2003
  2. jcsd
  3. Sep 4, 2003 #2
    I think terminal velocity for a human body is around 90 m/s, but don't hold me to that. So he would accelerate to that point and then presumably move at a constant velocity for however long. Assuming the acceleration bleeds away at a constant rate, you can just use s=.5at^2 + jt^3/6 for the distance travelled while he was accelerating and x=vt + s for the distance travelled once he reaches terminal velocity.

    At least, that's how I'd approach it.
  4. Sep 4, 2003 #3
    i'm sorry. what does the "j" stand for?
  5. Sep 4, 2003 #4
    I remember reading somewhere that a human falling with minimum air resistance (it assumed that you like were doing the pencil dive you used to use in the pool) falls at 180 MPH, which comes out to 83.33 M / S.
  6. Sep 5, 2003 #5
    j was just my variable for jerk. The rate of change of the acceleration.
    Last edited: Sep 5, 2003
  7. Sep 5, 2003 #6


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    Did you miss word this? Jerk is rate of change of acceleration.

    Acceleration is rate of change of velocity.
  8. Sep 5, 2003 #7


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    Er... I don't think the acceleration does reduce in a constant way...

    Rather, in general we consider there to be a drag force to be proportional to velocity, making the acceleration inversely proportional to velocity. This means you are looking at exponential velocity graphs...


    d^2x/dt^2 = g - k * dx/dt where k is some constant
  9. Sep 5, 2003 #8
    Yeah, I did.
  10. Sep 7, 2003 #9


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    A falling body has air resistance well into the turbulent regime, where resistance force is proportional to the square of the speed. So the equation of motion is


    a = acceleration = dv/dt = d^2z/dt^2, g = acceleration of gravity, m = mass. Rather than k to denote the drag resistance parameter, we simplify the math by using the terminal speed V.


    We then get


    integrate twice and get the following.

    v = V tanh (gt/V)

    z = (V^2/g) log (cosh (gt/V))

    Let's take V = 90 m/s, g = 10 m/s/s. Then

    v = 90m/s tanh(t/9sec)
    z = 810m (log(cosh(t/9sec))

    t(sec) v(m/s) z(m)
    0 0. 0.
    6 52.45 168.
    12 78.31 573.
    24 89.14 1602.
    30 89.77 2140.
    36 89.940 2679.
    42 89.984 3219.
    48 89.9958 3759.
    54 89.9989 4299.
    60 89.9997 4839.
  11. Sep 8, 2003 #10
    Krab, you're awesome! What level of physics is that--just out of curiosity?
  12. Sep 8, 2003 #11


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    You're welcome.

    The physics of constant force (no air resistance) is taught in 1st year or even in highschool.

    The physics of turbulence is something I learned on my own because I wanted to be able to calculate the performance of drag racers. But I guess some physics or engineering departments teach it in 2nd year.

    The solution of the equations requires math at the level of at least 2nd year.
  13. Sep 8, 2003 #12

    Chi Meson

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    "Differential equations" or "Diff-E-Q's." I've forgotten everything about them.
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