I am studying differential equations in university and I came across this problem:
A parachutist whose mass is $75$ kg drops from a helicopter hovering $4000\hbox{m}$ above the ground and falls towards the earth under the influence of gravity. Assume that the gravitational force remains constant during the flight, and that the drag due to air resistance is proportional to the velocity of the parachutist, with a proportionality constant $b = 150$ $N\hbox{s}/\hbox{m}$ when the chute is open. The chute opens immediately after the parachutist leaves the helicopter
The equation of motion (Newton’s 2nd law) they provide this:
$m \frac{dV}{dt} = mg − b\times V$
$m = 75$
$b = 150$
$g = 9.81$
The question: Obtain an expression for the velocity of the parachutist as a function of time. What is her/his terminal velocity? ($g = 9.81 \hbox{m}/\hbox{s}^2$)
I used the method of integrating factor to express the velocity as a function of time
$V = -4.905 + C e^{-2t}$ I think this answer the first part
I am having problem with the terminal velocity. I know it cannot be higher than $9.81,$ but I don't know what I should do to find this. My guess is either $4.905 \hbox{m}/\hbox{s}^2$ or $9.81$.