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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!

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

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  • $\begingroup$ I think I understand what you mean, but you could you provide some assistance in getting me started or in reference to this problem? I'm still confused $\endgroup$ – Billy Thorton Sep 15 '15 at 20:40
  • $\begingroup$ But also it is a separable equation, so you can solve it as such ;) $\endgroup$ – Evgeny Sep 15 '15 at 20:44
  • $\begingroup$ Sure, there are many ways. $\endgroup$ – Ivan Neretin Sep 15 '15 at 20:46
  • $\begingroup$ I recognize that it is separable, but as mentioned in the question, I'm not sure exactly what to do. Do you think you could give me more guidance on how to go about solving the problem? I don't really understand everything in the link provided $\endgroup$ – Billy Thorton Sep 16 '15 at 2:09
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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.

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  • $\begingroup$ So unlike some of the other answers, you say use the second equation I derived with $ds/dt$? How do I find the terminal velocity then? Perhaps I'm still a little confused? $\endgroup$ – Billy Thorton Sep 15 '15 at 20:44
  • $\begingroup$ In the velocity term coefficient of $ e^{-k t }$ sort of term. $\endgroup$ – Narasimham Sep 15 '15 at 21:05
  • $\begingroup$ How can I find terminal velocity from that term? $\endgroup$ – Billy Thorton Sep 16 '15 at 2:11
  • $\begingroup$ After finding $s$ expression as a function of $t$, finding terminal velocity = $ds/dt$ as $ t \rightarrow \infty $ should be no problem. $\endgroup$ – Narasimham Sep 16 '15 at 7:06
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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.

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  • $\begingroup$ I'm not sure I understand why you multiplied by the exponential. And in this particular solution, what would Ay(x) and B(x) be? $\endgroup$ – Billy Thorton Sep 15 '15 at 20:37
  • $\begingroup$ Multiplying by an exponential is done to make the LHS look like a product rule. In the example started with, A= c/m and B= -g. $\endgroup$ – Mark Sep 28 '15 at 20:58

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