Results 1 to 3 of 3

Math Help - The Parachute Problem

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    1

    Exclamation The Parachute Problem

    A parachutist falls from a balloon at a height of y0 (=4000). k1 represents air resistance in stage one and k2 in stage 2. Air resistance is assumed proportional to velocity.

    I've split this problem into two sections - free fall and after the parachute is deployed.

    For the first section the initial conditions are v(0) = 0 and I have the formula

    v(t) = (mg/k)(e^{-k1t/m} - 1)

    and for stage two I have

    v(t) = (mg/k1)(e^{-k1td/m} - 1)(e^{(-k2/m)(t-td)}) + (mg/k2)(e^{(-k2/m)(t-td)} - 1)

    where td is the time of deploying the parachute.

    I've integrated the first stage v(t) to find y(t), but I'm struggling to integrate the second stage. Please help.
    Last edited by silverstar45; February 25th 2008 at 12:44 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,931
    Thanks
    334
    Awards
    1
    Quote Originally Posted by silverstar45 View Post
    A parachutist falls from a balloon at a height of y0 (=4000). k1 represents air resistance in stage one and k2 in stage 2. Air resistance is assumed proportional to velocity.

    I've split this problem into two sections - free fall and after the parachute is deployed.

    For the first section the initial conditions are v(0) = 0 and I have the formula

    v(t) = (mg/k)(e^[-k1t/m] - 1)

    and for stage two I have

    v(t) = (mg/k1)(e^[-k1td/m] - 1)(e^[(-k2/m)(t-td)]) + (mg/k2)(e^[(-k2/m)(t-td)] - 1)

    where td is the time of deploying the parachute.

    I've integrated the first stage v(t) to find x(t), but I'm struggling to integrate the second stage. Please help.
    If someone doesn't get to it before I get back online this evening, then I will take a crack at it.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,931
    Thanks
    334
    Awards
    1
    Quote Originally Posted by silverstar45 View Post
    A parachutist falls from a balloon at a height of y0 (=4000). k1 represents air resistance in stage one and k2 in stage 2. Air resistance is assumed proportional to velocity.

    I've split this problem into two sections - free fall and after the parachute is deployed.

    For the first section the initial conditions are v(0) = 0 and I have the formula

    v(t) = (mg/k)(e^[-k1t/m] - 1)

    and for stage two I have

    v(t) = (mg/k1)(e^[-k1td/m] - 1)(e^[(-k2/m)(t-td)]) + (mg/k2)(e^[(-k2/m)(t-td)] - 1)

    where td is the time of deploying the parachute.

    I've integrated the first stage v(t) to find y(t), but I'm struggling to integrate the second stage. Please help.
    Letting t_d be the time of deployment:
    v(t) = (mg/k)(e^{-k_1t/m} - 1); 0 \leq t < t_d

    This is correct.

    After t = t_d we have a new drag coefficient k_2:
    v(t) = (mg/k)(e^{-k_2(t - t_d)/m} - 1); t_d \leq t

    I'm not sure where you got your second equation. You should be able to integrate these fairly easily. Notice that there will be a discontinuity in the velocity function, but the displacement needs to be continuous, so that gives you one boundary condition.

    If you need more help, just ask.

    (By the way, check out my LaTeX coding. You had most of it right; { } not [ ] groups symbols on the forum.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Parachute problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 24th 2008, 12:05 PM

Search Tags


/mathhelpforum @mathhelpforum