# A woman bails out of an airplane...

• Sep 21st 2011, 03:46 AM
alexmahone
A woman bails out of an airplane...
A woman bails out of an airplane at an altitude of 10,000 ft, falls freely for 20 s, then opens her parachute. How long will it take her to reach the ground? Assume linear air resistance $\displaystyle \rho v\ ft/s^2$, taking $\displaystyle \rho=0.15$ without the parachute and $\displaystyle \rho=1.5$ with the parachute. (Suggestion: First determine her height above the ground and velocity when the parachute opens.)

Without the parachute:

$\displaystyle \frac{dv}{dt}=32-0.15v$

$\displaystyle \int \frac{dv}{32-0.15v}=\int dt$

$\displaystyle \frac{-1}{0.15}ln\ (32-0.15v)=t+C_1$

$\displaystyle v(0)=0$

$\displaystyle C_1=\frac{-1}{0.15}ln\ 32$

$\displaystyle \frac{-1}{0.15}ln\ (32-0.15v)=t-\frac{1}{0.15}ln\ 32$

When $\displaystyle t=20$,

$\displaystyle \frac{-1}{0.15}ln\ (32-0.15v_{20})=20-\frac{1}{0.15}ln\ 32$

$\displaystyle -ln\ (32-0.15v_{20})=3-ln\ 32=-0.466$

$\displaystyle 32-0.15v_{20}=1.593$

$\displaystyle v_{20}=202.712$

$\displaystyle -ln\ (32-0.15v)=0.15t-ln\ 32$

$\displaystyle ln\ \left(\frac{32}{32-0.15v}\right)=0.15t$

$\displaystyle \left(\frac{32}{32-0.15v}\right)=e^{0.15t}$

$\displaystyle 32-0.15v=32e^{-0.15t}$

$\displaystyle 0.15\frac{dx}{dt}=32(1-e^{-0.15t})$

$\displaystyle \int \frac{3}{640}dx=\int (1-e^{-0.15t})dt$

$\displaystyle \frac{3x}{640}=t+\frac{e^{-0.15t}}{0.15}+C_2$

$\displaystyle x(0)=0$

$\displaystyle C_2+\frac{1}{0.15}=0$

$\displaystyle C_2=-\frac{1}{0.15}$

$\displaystyle \frac{3x}{640}=t+\frac{e^{-0.15t}}{0.15}-\frac{1}{0.15}$

When $\displaystyle t=20$,

$\displaystyle \frac{3x_{20}}{640}=20+\frac{e^{-3}}{0.15}-\frac{1}{0.15}$

$\displaystyle x_{20}=2915.253$

With the parachute:

$\displaystyle \frac{dv}{dt}=32-1.5v$

$\displaystyle \frac{-1}{1.5}ln\ |32-1.5v|=t+C_3$

$\displaystyle ln\ |32-1.5v|=-1.5(t+C_3)$

$\displaystyle 32-1.5v=\pm e^{-1.5C_3}e^{-1.5t}=Be^{-1.5t}$ [where $\displaystyle B=\pm e^{-1.5C_3}$].

$\displaystyle v(20)=202.712$

$\displaystyle -272.068=B*e^{-30}$

$\displaystyle B=-2.907*10^{15}$

$\displaystyle 32-1.5v=-2.907*10^{15}e^{-1.5t}$

$\displaystyle 32-1.5\frac{dx}{dt}=-2.907*10^{15}e^{-1.5t}$

$\displaystyle 1.5\frac{dx}{dt}=32+2.907*10^{15}e^{-1.5t}$

$\displaystyle \int 1.5dx=\int (32+2.907*10^{15}e^{-1.5t})dt$

$\displaystyle 1.5x=32t-1.938*10^{15}e^{-1.5t}+C_4$

$\displaystyle x(20)=2915.253$

$\displaystyle 4372.88=640-1.938*10^{15}e^{-30}+C_4$

$\displaystyle C_4=3914.231$

$\displaystyle 1.5x=32t+1.938*10^{15}e^{-1.5t}+3914.231$

When $\displaystyle x=10,000$,

$\displaystyle 15000=32t+1.938*10^{15}e^{-1.5t}+3914.231$

$\displaystyle 32t+1.938*10^{15}e^{-1.5t}=11085.769$

$\displaystyle t\approx 346\ s=5\ min\ 46\ s$
• Sep 21st 2011, 05:23 AM
CaptainBlack
Re: A woman bails out of an airplane...
It's a pity that you have had this set with an air resistance model appropriate for low Reynolds number motion in a situation where the Reynolds number is high.

CB