Answer to Question #121092 in Differential Equations for Joseph Se

Question

1. A skydiver weighing 174 lbf (including equipment) falls vertically downward from an altitude of 4000 ft and opens the parachute after 10s of free fall. Assume that the force of air resistance, which is directed opposite to the velocity, is 0.76 |v| when the parachute is closed and 12|v| when the parachute is open, where the velocity v is measured in ft/s.

Measure the positive direction of motion downward. Consider that x(0) = 0 ft. Use g = 32 ft/s^2. Round your answers to two decimal places.

(a) Find the speed of the skydiver when the parachute opens. (the tolerance is +/- 2 percent)
v(10) = _______ ft/s

(b) Find the distance fallen before the parachute opens.(the tolerance is +/- 2 percent)
x(10) = _________ft

(c) What is the limiting velocity vl after the parachute opens?(the tolerance is +/- 2 percent)
vl = _____ ft/s

Answer 1

a)

$m\frac {dv}{dt}=mg-bv$

where b is air resistance

Let:

$k=b/m$

then:

$dv/dt+kv=g$

First, solve:

$dv/dt+kv=0$

$dv/dt=-kv$

$dv/v=-kdt$

$v=Ce^{-kt}$

Then:

$\frac {d}{dt}C(t)=ge^{kt}$

$v(t)=e^{-kt}(v(0)-g/k)+g/k$

Since $v(0)=0$ :

$v(t)=\frac {mg}{b}(1-e^{-bt/m})$

Answer:

$v(10)=v(10)=v(10)= \frac {174\cdot32}{0.76}(1-e^{-0.76\cdot10/174})=313.11$ ft/s

b)

$v(t)=dx/dt$

$x(t)=\int v(t)dt=\frac {mg}{b}(t+\frac {m}{b}e^{-bt/m})+C$

$x(0)=0=\frac {m^2g}{b^2}+C$

$C=-\frac {m^2g}{b^2}$

$x(t)=\frac {mg}{b}(t+\frac {m}{b}e^{-bt/m})-\frac {m^2g}{b^2}$

Answer:

$x(10)=\frac {174\cdot32}{0.76}(10+\frac {174}{0.76}e^{-0.76\cdot10/174})-\frac {174^2\cdot32}{0.76^2}=1576.96$ ft

c)

$v(t\to\infin)=mg/b=174\cdot32/12=464$ ft/s