[SOLVED] 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 $\rho v\ ft/s^2$ , 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 $x=10,000$ ,

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

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

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