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Diffy Q word Problem ?

  1. Jun 1, 2008 #1
    1. The problem statement, all variables and given/known data

    A person bails out of an airplane at an a;titude of 15,000 feet, falls freely for 35 seconds, then opens the parachute. How long will it take the person to reach the ground? Assume linear air resistance pV ft/s^2, paking p=.17 without the parachute and p=1.5 with the parachute.

    2. Relevant equations

    3. The attempt at a solution
    v(0) = 0
    The text book is very vague and unsure how to even approach the set up of this equation please help with detailed explaination. Thank you so very much.

  2. jcsd
  3. Jun 1, 2008 #2


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    Welcome to PF!

    Hi LadyAnn ! Welcome to PF! :smile:

    The acceleration downwards is g minus pv.

    And acceleration = dv/dt.

    So solve for v as a function of t. :smile:
  4. Jun 1, 2008 #3
    Hi LadyAnn

    The way I usually tackle these problems is to start with the expression [tex]F = m\frac{d^{2}s}{dt^{2}}[/tex] such that [tex]\frac{d^{2}s}{dt^{2}} = F/m[/tex]
    now m doesn't appear in the information you gave so I'd be tempted to assume that F = mg - [tex]\hat{p}\frac{ds}{dt}[/tex] where [tex]\hat{p} = mp[/tex] (ie: I'm assuming p is a lumped constant)

    With this you can reduce it to a second order ODE of the form:
    [tex]\frac{d^{2}s}{dt^{2}}+p\frac{ds}{dt} = g[/tex] and then solve for s using the fact that at t = 0 you know how fast the person is falling and where the person is.
    You can then use the other information to find the required time to fall 15,000ft
    Last edited: Jun 1, 2008
  5. Jun 1, 2008 #4
    don't you complicate things?

    Mine method:
    Treat this as a physics problem. Use those motion formulas
    Part 1 - free fall
    Part 2 - parachute opens

    Find distance covered in part 1: say D1

    Distance left = 15,000 - D1
    now using acceleration = g - p = a
    find final velocity after D1
    find time for remaining distance using one of fundamental motion formulas...
    use d = vt + 0.5at^2 one...
  6. Jun 1, 2008 #5
    Hmm...I might be wrong but when you say that a = g-p I infer that the acceleration is constant...surely this is not true though because it depends on how fast that person is falling at any time (air resistance)...and so using the suvat equations would not be correct?

    The reason I would reduce it to a second order ODE in terms of position (as opposed to first order in terms of velocity) is that position is surely required to solve for t once the parachute opens?
    Last edited: Jun 1, 2008
  7. Jun 1, 2008 #6
    yep, you seem to be right..
    there's "v" after p and I missed it...
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