The volume of a sphere:

$\overline{){\mathbf{V}}{\mathbf{=}}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\pi}}{{\mathbf{r}}}^{{\mathbf{3}}}}$

Terminal velocity:

$\overline{){\mathbf{v}}{\mathbf{=}}\sqrt{\frac{\mathbf{2}\mathbf{m}\mathbf{g}}{\mathbf{\mu}\mathbf{\rho}\mathbf{A}}}}$

Density:

$\overline{){\mathbf{\rho}}{\mathbf{=}}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}{\mathbf{v}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}}}$

The diameter of raindrop, d = 3.2 mm (1m/1000mm) = 0.0032 m

r = d/2 = 0.0032/2 = 0.0016 m

mass, m = ρ_{water}•V = 1000 × (4/3)π × (0.0016)^{3} = 1.716 × 10^{-5} kg

Cross sectional area, A = πr^{2} = π(0.0016)^{2} = 8.042 × 10^{-6} m^{2}

Calculate the velocity of a spherical raindrop falling from 5.4 km.

Take the size across the rain drop to be 3.2 mm, the density of air to be 1.25 kg/m^{3}, density of water to be 1000 kg/m^{3}, the surface area to be πr^{2}, and the drag coefficient to be 1.0.

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