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Im stuck, modeling problem

  1. Jul 13, 2009 #1
    A sky diver weighing 180 lb falls vertically downward from an altitude of 5000ft and opens the parachute after 10 s of free fall. assume that the force of air resistance is 0.75(v) when the parachute is closed and 12(v) when the parachute is open, where the velocity v is measured in ft/s.

    a) find speed of the sky diver when the parachute opens
    b) find the distance fallen before the parachute opens
    c)What is the limiting velocity V_L after the parachute opens?
    d) Determine how long the sky diver is in the air after the parachute opens

    I have tried to set it up like -

    (dv/dt)= (.75(v))/180-a and seperating and solving but answer doesnt make sense

    and

    V=32(t)-(.75v/180)t that didnt make sense either

    I know I am probably setting this up wrong, I was just trying to do F=ma for the set up? please point me in the right direction. Thanks!
     
  2. jcsd
  3. Jul 13, 2009 #2

    HallsofIvy

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    I have no idea where you got that first equation- it has "180- a" and there is no reference to "a" in what you write before that. The second equation is almost right but you seem to be taking the positive direction of velocity downward which seems peculiar to me. More importantly, you have the acceleration on the right side equal to V rather than dV/dt. Finally, 180 pounds is weight not mass. Weight is the force due to gravity on a given mass: 180= mg so m= 180/g.

    dv/dt= acceleration= force/mass and there are two forces acting here: gravitational force -mg and air resistance -(.75)v (- because the resistance is always directed opposite to the velocity it is resisting.)

    m dv/dt= -mg- (.75)v so dv/dt= -g- (.75)v/m= -g- (.75)gv/180= -g(1+ (.74/180)v).

    That is an easily separable equation. Solve for v(t) and determine v(10), the speed when the parachute opens. After the parachute opens the equation is
    dv/dt= -g(1+ (12/180)v). Solve that using the previous v(10) as the initial value.

    Hopefully, you will find that v(t) is an exponential with negative exponent. The "limiting velocity" is the limit as t goes to infinity. Even simpler, at the "limiting velocity", the velocity does not change so dv/dt= -g(1+ (12/180)v)= 0.
     
  4. Jul 14, 2009 #3
    I was using a to be acceleration or dv/dt. But thankyou I knew I wasnt seeing it right.
     
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