Question

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

According to the question,

(a) Equation of the motion before parachute open,

$ma = mg - 0.76v$

Equation of motion after parachute open,

$ma' = mg - 12v'$

Now, For motion before parachute open,

$m\frac{dv}{dt} = mg - 0.76v$

$\int \frac{mdv}{(mg-0.76v)} = \int dt$

Integrating both sides,

$\frac{-m}{0.76} ln|{mg-0.76}| = t + C$

applying condition that $v (0) = 0$

$C = \frac{-m}{0.76} ln|mg|$

equation of the motion is given by,

$\frac{-m}{0.76} ln|{mg-0.76}| = t + \frac{-m}{0.76} ln|{mg}|$

$\frac{-m}{0.76} ln|\frac{mg-0.76}{mg}| = t$

or $v = \frac{mg}{0.76}(1-e^{-\frac{0.76t}{m}})$

so velocity when parachute open,

putting values, mg = 174*32, m = 174 lb,

$v= \frac{174*32}{0.76} (1-e^{-\frac{0.76*10}{174}}) =306.4$ ft/s

(b) for distance calculation,

$\frac{dx}{dt} = \frac{mg}{0.76}(1-e^{-\frac{0.76t}{m}})$

$\int_0^x {dx} = \int_0^{10} \frac{mg}{0.76}(1-e^{-\frac{0.76t}{m}}){dt}$

on integrating, we get,

$x = \frac{mg}{0.76}(t+\frac{m}{0.76}e^{-\frac{0.76t}{m}})|_0^{10}$

putting values, we get

$x = \frac{174*32}{0.76}(10+\frac{174}{0.76}e^{-\frac{0.76*10}{174}})= 1556.8 ft$

(c) for critical value,

$\frac{dv}{dt} = 0$

so we obtain,

$mg = 0.76v$

$\implies v = \frac{174*32}{0.76} = 7326.3$ ft/s