A woman bails out of an airplane...

**alexmahone** · September 21st 2011, 03:46 AM

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 $\rho v\ ft/s^2$ , taking $\rho=0.15$ without the parachute and $\rho=1.5$ with the parachute. (Suggestion: First determine her height above the ground and velocity when the parachute opens.)

Without the parachute:

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

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

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

$v(0)=0$

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

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

When $t=20$ ,

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

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

$32-0.15v_{20}=1.593$

$v_{20}=202.712$

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

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

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

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

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

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

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

$x(0)=0$

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

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

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

When $t=20$ ,

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

$x_{20}=2915.253$

With the parachute:

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

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

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

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

$v(20)=202.712$

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

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

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

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

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

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

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

$x(20)=2915.253$

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

$C_4=3914.231$

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

When $x=10,000$ ,

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

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

$t\approx 346\ s=5\ min\ 46\ s$

**CaptainBlack** · September 21st 2011, 05:23 AM

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

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