Differential Equations Skydiver Problem

Question

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!

Answer 1

That's a linear ODE with constant coefficients, and should be solved as such.

Answer 2

This is the standard second order linear ODE. Directly solvable without using integrating factor. So I omit details. $ds/dt$ tends to a terminal velocity limit by virtue of negative exponent.

Answer 3

Don't make it harder by making the change of variable from velocity to position. While it is inherently the same problem, as stated you have a first order differential equation which can always (almost always?) be solved by integration.

For example, take the completely different and in no way similar ODE: $$\frac{dy(x)}{dx}+Ay(x)= B(x)$$

Then a simple trick to solving this is to multiply by the exponential $e^ {\int A dx}$

$$e^ {\int A dx} \frac{dy(x)}{dx}+A e^ {\int A dx} y(x)= B(x) e^ {\int A dx}.$$

The left hand side can be recognized as the product rule such that $$\frac{d}{dx}(y(x)e^ {A x})= B(x) e^ {A x}$$ Using seperation of variables: $$=d(y(x)e^ {A x})= B(x) e^ {A x} dx$$ $$=\int d(y(x)e^ {A x})= \int B(x) e^ {A x} dx$$ And this leaves you with a solution of the form $$y(x)= Ce^{-Ax}+e^{-Ax}\int B(x) e^{Ax}dx$$

Here C is your constant of integration that helps you match boundary/initial conditions.

This method can help you find the velocity in the problem above, and once you have v(t) you then know how to find the solution for position as you said $\frac{ds}{dt}=v(t)$.

Hope this helps.