Back to the N=1 (Linear Drag) case.

4. Quadratic (N=2 case) drag

The more typical quadratic drag case has C = K₂/m and a = V_max = $\sqrt{mg/K_2}$, so the integration problem becomes

$$\int_{0}^{v} \frac{dv}{a^2 - v^2} \; = \; C t .$$ (9)

We identify the left side as the standard integral that gives the inverse hyperbolic tangent:

$$\int_{0}^{v} \frac{dv}{a^2 - v^2} \; = \; \frac{1}{a} \tanh^{-1}\left(\frac{v}{a}\right) \biggr\vert_{0}^{v}$$

which gives

$$\frac{1}{a} \tanh^{-1}\left(\frac{v}{a}\right) \; = \; Ct .$$

[As an aside, we note that this can also be done with logs.]

If we define B = aC = $\sqrt{gK_2/m}$ and take the tanh of both sides of the equation, we can rewrite this as the solution for v(t):

$$v(t) \; = \; V_{\rm max} \tanh(Bt)$$ (10)

where V_max = $\sqrt{mg/K_2}$ and we note that B has the role of the inverse of a time constant in the hyberbolic tangent function. (Define T₂ = $\sqrt{m/gK_2}$ = 1/B as that time constant.)

Further, if we assume that the initial position is y(0) = 0, remember that the positive direction for y is down, and integrate the velocity with respect to time, we get the result

$$y(t) \; = \; \left( \frac{V_{\rm max}}{B} \right) \ln [ \cosh (Bt) ]$$ (11)

and we can solve H = y(T) for T to find the time to fall a distance H.

The validity of this solution can be checked by explicit differentiation with respect to t and is left as an exercise.

Next, Sample Results.