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Air drag is approximated by the following equation

$$F=\frac12 \rho A v^2 C_D$$

Depending on the drag coefficient $C_D$, this force may be linear or quadratic with respect to velocity $v$. If $v$ is large, it is quadratic and if $v$ is small, it is linear.

Raindrops begin falling with velocity $0$ and accelerate subject to gravitational acceleration and linear drag. When it reaches terminal velocity, is the drag force on the raindrop quadratic or linear?

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Drag force nature depends not exactly on object velocity, but rather on Reynolds number, which is : $$ \mathrm {Re} ={\frac {uL}{\nu }} $$ where $\nu$ is kinematic viscosity of fluid, $u$ - fluid flow speed around object (object may be stationary with respect to the ground, but not with respect to the fluid flow) and $L$ is cross-section diameter of object. Reynolds number is crucial in determining drag force nature, because same drag force can be generated on low velocity + high cross-section objects as for high velocity + low cross-section objects.

If object is spherical-like (which is the case of raindrops) AND Reynolds number is small, namely $Re \ll 1$, then drag force will be Stokes' law, which was exactly extrapolated solving Navier–Stokes equations for small Reynolds numbers :

$$ F_{\rm {d}}=6\pi \mu R~v $$

Otherwise full-scale drag force expression (quadratic form) must be used : $$ F_{d}\,=\,{\tfrac {1}{2}}\,C_{D}\,A\,\rho ~v^{2} $$

So one needs to determine Reynolds numbers for a raindrop, to answer your question. Assuming air kinematic viscosity of $1.48 × 10^{−5}~m^2/s$, and depending on raindrop terminal velocities and diameters, one can calculate Reynolds numbers:

  • Drizzle drops (Ø 0.2 - 0.5 mm), $Re \in [9;67]$
  • Rain drops (Ø 0.6 - 4 mm), $Re \in [81;2432]$
  • Large drops (>Ø 5 mm), $Re \gt 4392$

In all three cases Reynolds number is NOT a small value, i.e. $Re \ll 1$ condition is not satisfied, thus quadratic form of drag force must be used. Btw, also drag force nature can be determined empirically.

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