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Parachutist freefalling question

  1. May 15, 2012 #1
    http://dl.dropbox.com/u/33103477/parachutist.png [Broken]

    (This is from a mathematics not physics exam)

    Well what I've done, is to compose the equation describing the velocity of the parachute once it has opened. I did this by setting up the differential equation as:

    [tex] \frac{dv}{dt} = g-kv [/tex]

    I solved that to get:

    [tex] v = \frac{g}{k} - \frac{Ce^{-kt}}{k} [/tex]

    Now the parachutist was travelling at v_0 when it deployed so I can say V(0)=v_0

    hence solving I got:

    [tex] v(t)= \frac{g}{k} - \frac{g-v_0k}{k}e^{-kt} [/tex]

    Now the point is If I know how far from the ground the parachute opened I can figure out the distance travelled since the parachute was opened ?

    I'm confused so any direction on how I should get to x(t) from v(t) would be useful.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 16, 2012 #2

    ehild

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    Homework Helper
    Gold Member

    The distance travelled is the integral of v(t). (V, the velocity, is the derivative of the displacement v=dx/dt. The parachutist travels vertically downward, so the displacement is the same as the distance travelled here.)

    You made a small error in the first equation: dv/dt=-(k/m)v.

    ehild
     
    Last edited: May 16, 2012
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