I've seen many variants of this problem online, but not quite the same as this, so I don't believe this is a duplicate.
The famous differential equation problem models a skydiver jumping out of a plane as
$$m \frac{dv}{dt} = -mg-cv$$
where m is the mass of the skydiver, g is acceleration due to gravity (9.81 m/s) and c is the coefficient of air resistance.
Here is my question:
Suppose the skydiver exits the stationary helicopter at an altitude of 2000m and opens the parachute at 500m. For this exercise, the parachute opens instantaneously. Suppose m= 70kgs, the coefficient of air resistance is c = 14 for a free falling body and c = 105 after the parachute opens. When does the skydiver open the parachute? When does the Skydiver hit the ground?
So my attempt at a solution kinda starts like this, but this is where I get stuck:
Let's assume that s(t) is the distance the body falls in time t from its initial point of release, then $v = \frac{ds}{dt}$ and $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$. Thus we have a second order differential equation: $m \frac{d^2s}{dt^2} = -mg-c\frac{ds}{dt}$. By simply manipulation, we get $m \frac{d^2s}{dt^2} + c\frac{ds}{dt} = mg$. I originally started thinking of finding the integrating factor, however, this seems more complicated and won't help produce the equation that I need.
Would anyone be able to help? Much thanks in advanced!
Best Answer
That's a linear ODE with constant coefficients, and should be solved as such.