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Differential Equation Question

  1. Jun 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Felix, a Red bull skydiver, (mass m) who drops out of a hovering helicopter, falls long enough without a parachute(so the drag force has strenght kv2 ) to reach his first terminal velocity ( denoted v1). When his parachute opens, the air resistance force has a strenght Kv.

    At what minimum altitude must his parachute open so that he slows to within 1% of his new
    (much lower) terminal velocity (v2) by the time he hits the ground?

    2. Relevant equations

    newtons second law F = ma

    3. The attempt at a solution
    I used newton second law of motion and think i have to use the time t = 0 and not sure what to do again could someone help me
     
  2. jcsd
  3. Jun 21, 2013 #2

    Mark44

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    Please show us what you did and we can go from there.
     
  4. Jun 23, 2013 #3
    F=ma
    kv=md2x/dt2

    kdx/dt=md2x/dt2

    from here i don't know where to go i don't even know if this is correct does any one know how to work it out ???
     
  5. Jun 23, 2013 #4
    This is what i did
     
  6. Jun 23, 2013 #5

    tiny-tim

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    what about gravity? :redface:
     
  7. Jun 23, 2013 #6
    i don't know i think the gravity will be the acceleration:confused:
     
  8. Jun 23, 2013 #7

    tiny-tim

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    gravity (times mass) is one of the forces on the skydiver

    kv is the other force

    now use Ftotal = ma :smile:
     
  9. Jun 23, 2013 #8
    ok will try something:approve:
     
  10. Jun 23, 2013 #9

    haruspex

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    Just checking: it's kv2 without parachute but Kv with? Not Kv2?
     
  11. Jun 24, 2013 #10
    yes when the parachute opens it is Kv and before it is Kv2 i am mostly sure we have o use the Kv since the question asked when the parachute was open
     
  12. Jun 24, 2013 #11

    haruspex

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    That's not quite what it says in the OP. There you wrote kv2, not Kv2. I can cope with the change from v2 to v as the parachute opens, but it seems very unlikely the coefficient would be the same. For a start, its dimensions would be different.
    However, since you're given the first terminal velocity as v1, it's irrelevant what the drag was before the parachute opened. So ignore my questions and concentrate on getting the equation when drag = Kv.
     
  13. Jun 26, 2013 #12
    this is what i did so far

    ma = mg - kv (since kv is the opposite direction as the gravitational force)

    md2x/dt2 = mg - k dx/dt
     
  14. Jun 26, 2013 #13
    this was given

    Mass of Felix = 70 kg
    Gravity = 9.8ms^-2
    Air resistance proportional constant K = 110 kgs^-1
    terminal velocity without parachute, v1 = 52ms^-1
    Terminal velocity with parachute v2 = 6.2ms^-1

    what do i do from here ???????????
     
  15. Jun 26, 2013 #14

    tiny-tim

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    ok, now rewrite that as mdv/dt = mg - kv, and integrate …

    what do you get? :smile:
     
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