When an object is in free fall, we have:

$$a(t) = g - \frac{c}{m}v(t)^2$$

where $g$ is acceleration due to gravity, $m$ is the mass of the object, and $c$ is the coefficient of air resistance.

How does one get the distance traveled after t seconds? I tried integrating it, giving

$$v(t) = gt - \frac{c}{3m}s(t)^3$$

$$s(t) = \sqrt[3]{\frac{3m}{c}(gt - v(t))}$$

Which is a function defined in terms of its derivative. How would I find a better definition of $s(t)$?


3 Answers 3


You're right about the equation of motion for an object in free fall with air resistance (well, almost right: your $c$ is not the usual definition of the drag coefficient), but when you integrate it, you don't go from $v^2$ to $s^3/3$. That only works when the thing being squared is actually the variable of integration: $\int t^2\mathrm{d}t = t^3/3$, but $\int f(t)^2\mathrm{d}t \neq f(t)^3/3$.

To properly solve the equation, you'll need to start by finding speed as a function of time. You can write the equation as

$$\frac{\mathrm{d}v}{\mathrm{d}t} = g - \frac{c}{m}v^2$$

This is a separable differential equation, so you can put everything involving the independent variable $t$ on one side and everything involving the dependent variable $v$ on the other side,

$$\frac{\mathrm{d}v}{g - \frac{c}{m}v^2} = \mathrm{d}t$$

This can be integrated over $t$, giving

$$t = \int_{v(0)}^{v(t)}\frac{\mathrm{d}v}{g - \frac{c}{m}v^2} = \sqrt{\frac{m}{cg}}\tanh^{-1}\biggl(\sqrt{\frac{c}{mg}}v\biggr)$$

(assuming $v(0) = 0$). Then you can solve this for velocity,

$$\frac{\mathrm{d}s}{\mathrm{d}t} = v = \sqrt{\frac{mg}{c}}\tanh\biggl(\sqrt{\frac{cg}{m}}t\biggr)$$

which is another separable equation,

$$\int_{s(0)}^{s(t)}\mathrm{d}s = \int_0^t\sqrt{\frac{mg}{c}}\tanh\biggl(\sqrt{\frac{cg}{m}}t\biggr)\mathrm{d}t$$

The result of that integration is

$$s(t) = s(0) + \frac{m}{c}\log\cosh\biggl(\sqrt{\frac{cg}{m}}t\biggr)$$

I've written a blog post about a (possibly) interesting "application" of this calculation. The math above is basically a summary of part of that post.

  • $\begingroup$ Thank you, that's perfect. Turns out that integral isn't as trivial as it first appears. $\endgroup$ Nov 9, 2010 at 23:37
  • 2
    $\begingroup$ I recently wrote a blog post about it, too! It turns out this situation has an interesting link with constant acceleration in special relativity. arcsecond.wordpress.com/2010/10/25/… $\endgroup$ Nov 10, 2010 at 5:59

The equation you're looking for is

$$m\frac{dv}{dt}=\frac{1}{2}\rho C_{D} A v^{2}-mg$$

where $C_{D}$ is the drag coefficient and $A$ is the cross-sectional area of the object.

For the vertical position as a function of time, you can see the solution in Free Fall.

By the way, your calculation is not quite right. You have $s(t)$ (that is, a function which depends on the time) but you're integrating as if $s(t)$ were $t$ (which is not necessarily true).


When accelereration is given as a function of velocity only then the following apply

$$ x(u)=\int\frac{u}{a(u)}\,\mathrm{d}u+K_{1} $$ and $$ t(u)=\int\frac{1}{a(u)}\,\mathrm{d}u+K_{2} $$

the rest is solved similarly to David Z (above).


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