*This article appeared originally in my Day Book.*

I was watching the *Mythbusters* today, and they were testing a myth that involved a human falling from from great height. They explained the concept of terminal velocity, and stated that the terminal velocity of a human body is approximately 120 mph. They then proceeded with a calculation, according to which it takes ~5.5 seconds for a falling body to reach this velocity.

Except that it takes almost twice as long.

120 mph is about 192 kph, or ~53.3 m/s. At an acceleration of 9.81 m/s^{2}, it indeed takes about 5.44 seconds to reach this velocity. That is, if you are falling *in a vacuum*.

But we're talking about falling in air, aren't we. There's no terminal velocity in a vacuum; a falling body would continue to accelerate until it hits the ground. In air, however, air resistance doesn't just kick in at the moment when you reach terminal velocity; it is there all the time, proportional to the square of your velocity, which is how it gets to be equal the to force of gravity eventually when you reach terminal velocity.

So how long would it take for a falling body to reach terminal velocity? Wrong question! A falling body never actually reaches terminal velocity; as air resistance increases, the acceleration of the body decreases, so it'd approach, but never quite get to, that terminal velocity.

So then... how long would it take for it to reach 95% of its terminal velocity?

The acceleration $a$ of a rapidly falling body has two components: a constant acceleration $g$ due to gravity, and a variable acceleration that is a function of the square of the velocity $v$, multiplied by a proportionality factor $\mu$ ($\mu=\rho AC_d/2m$ where $\rho$ is the air density, $A$ is the cross-sectional area, $C_d$ is the drag coefficient, and $m$ is the mass of the body):

\[a=g-\mu v^2.\]

But the acceleration $a$ is none other than the time derivative of the velocity, so we have a simple differential equation for the velocity as a function of time, i.e., $v(t)$:

\[\frac{dv}{dt}=g-\mu v^2.\]

Acceleration becomes zero when terminal velocity is reached. At this point, $g-\mu v_\mathrm{max}^2=0$, from which $\mu=g/v_\mathrm{max}^2$; if $g=9.81~{\rm m}/{\rm s}^2$ and $v_\mathrm{max}=53.3~\rm{m}/\rm{s}$, we get $\mu=0.0035$. (At $\rho=1.3~{\rm kg}/{\rm m}^3$, $m=80~{\rm kg}$, and $A=0.4~{\rm m}^2$, this means $C_d\simeq 1$.)

The solution to the differential equation comes in the form of

\[v=v_\mathrm{max}\tanh(t\sqrt{\mu g}),\]

and we now seek the value of $t$ for which

\[\tanh(t\sqrt{\mu g})=0.95.\]

The answer can be easily calculated:

\[t\simeq 10~{\rm s}.\]

Another question is, how much distance would the body travel in this amount of time? In the Mythbusters episode, they were using a figure of 500-odd feet, but that was again based on the same incorrect math (well, the math was correct, it was just applied incorrectly, so perhaps I should say, incorrect physics.) The distance traveled can be calculated by computing the integral of the velocity:

\[s=\int v_{\rm max}\tanh(t\sqrt{\mu g})~dt=\mu^{-1}\ln\cosh(t\sqrt{\mu g})\simeq 340~{\rm m},\]

which is over 1100 feet, or about twice the height that the Mythbusters have calculated.

But then, who cares about sloppy math (or physics!) when the episode provided an excuse to blow up some 500 pounds of high explosives? Fortunately, they did not make a mistake in *those* calculations, so nobody got hurt!