1. ## 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

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

and for stage two I have

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

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.

2. Originally Posted by silverstar45
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

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

and for stage two I have

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

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

3. Originally Posted by silverstar45
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

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

and for stage two I have

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

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 $\displaystyle t_d$ be the time of deployment:
$\displaystyle v(t) = (mg/k)(e^{-k_1t/m} - 1)$; $\displaystyle 0 \leq t < t_d$

This is correct.

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

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