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Wind resistance -- upwards acceleration, typically varies either linearly or quadratically by the current velocity.

There is a constant downward acceleration due to gravity.

How can we model the velocity over time of a falling object, subject only to wind resistance and downwards gravity?

I don't have much experience with differential equations, but I do know that this answer necessarily involves it, so could you possibly explain every step?

Thank you.

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would you mind rewording this question or making it community wiki (or both)? –  Tom Stephens Jul 27 '10 at 2:02
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1 Answer

up vote 2 down vote accepted

$$\begin{align*} \sum F &= ma\\ \frac{dv}{dt} &= a\\ &= \frac{\sum F}{m}\\ \sum F &= mg - kv\\ \frac{dv}{dt} &= g - \frac{k}{m} v\end{align*}$$

This is a differential equation with a solution of

$$\begin{align*} v &= A + B \space exp\left(\frac{-k}{m} t\right)\\ \frac{dv}{dt} &= - B \cdot \frac{k}{m} \cdot \exp\left(\frac{-k}{m} t\right) \end{align*}$$

Match terms and initial conditions ($v = 0$ at time $t = 0$) and you get $$\begin{align*} &g - \frac{k}{m}A - \frac{k}{m} B\cdot \exp\left(\frac{-k}{m} t\right)\\ \implies&-B \frac{k}{m} \cdot \exp\left(\frac{-k}{m} t\right)\\ \implies& A = g \cdot \frac{m}{k} \space \text{and} \space B = -g \cdot \frac{m}{k}\\ \implies& v = \frac{mg}{k} \cdot \left(1 - \exp\left(\frac{-k}{m} t\right)\right)\end{align*}$$

That's a linear differential equation ($\frac{dv}{dt}$ is a linear function of $v$); the $\sum F = -kv^2$ is a nonlinear differential equation (can't remember off the top of my head how to deal with that one; it may not have a closed form solution). It's a bit difficult to summarize the techniques in general, but any good book on differential equations would cover them.

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