# Where do the units of velocity come from when calculating velocity in free fall under a parachute?

I'm working on calculating descent velocity of an object under a parachute, but I'm running into issues when checking my units.

The formula for velocity with a parachute is (at least from my research) $$v = \sqrt {{{2w} \over {\rho {C_D}A}}}$$ In English units, $w$ is the weight of the object and parachute in $\rm{lb}$, $\rho$ is the air density in ${{{\rm{lb}}} \over {{\rm{f}}{{\rm{t}}^{\rm{3}}}}}$, $A$ is the area in $\rm{ft}^\rm{2}$, and $C_D$ is the unitless coefficient of drag.

Showing just the units, you get $$\sqrt {{{{\rm{lb}}} \over {{{{\rm{lb}}} \over {{\rm{f}}{{\rm{t}}^{\rm{3}}}}}{\rm{ \times f}}{{\rm{t}}^{\rm{2}}}}}} {\rm{ = }}\sqrt {{\rm{ft}}} \ne {{{\rm{ft}}} \over {{{\rm{s}}}}}$$

How does this work? Am I making a mistake somewhere?

• Is the density $\frac{lb}{ft^3}$ ? Is it not $\frac{lb*s^2}{ft^4}$? – Ben S Feb 6 '17 at 5:22

You're making a mistake, but it's a very subtle one (and indeed part of the reason most scientists don't like to use imperial units). There are actually two different units that go by the name of "pound". One is a unit of mass, and the other is a unit of force, sometimes designated $\mathrm{lbf.}$ when one wants to distinguish between them. They differ by a factor of $g$, the standard gravitational acceleration: $$\mathrm{lbf.} = g\times\mathrm{lb.}$$ Note that $g$ has units of distance per time squared. (It's about $32\ \mathrm{ft./s^2}$, but the specific value isn't important for my explanation.)
In your equation, the pound you use to measure the weight $w$ is the pound of force, but the pound you use to measure the air density is the pound of mass. If you're careful to maintain the distinction between these two, you'll see that the units work out correctly: $$\sqrt{\frac{\mathrm{lbf.}}{\frac{\mathrm{lb.}}{\mathrm{ft.}^3}\times\mathrm{ft.}^2}} \propto \sqrt{\frac{\mathrm{lb.}\times\mathrm{ft./s^2}}{\frac{\mathrm{lb.}}{\mathrm{ft.}^3}\times\mathrm{ft.}^2}} = \sqrt{\frac{\mathrm{ft.}^2}{\mathrm{s}^2}} = \frac{\mathrm{ft.}}{\mathrm{s}}$$