Terminal velocity of a falling object:

$\overline{){{\mathbf{v}}}_{{\mathbf{t}}}{\mathbf{=}}\sqrt{\frac{\mathbf{2}\mathbf{m}\mathbf{g}}{\mathbf{\rho}\mathbf{C}\mathbf{A}}}}$, where ρ is the density of the medium, C is the drag coefficient, A is the frontal area of the falling object, m is the mass of the object, and g is the acceleration due to gravity.

We'll also use the kinematic equation:

$\overline{)\mathbf{}{{{\mathbf{v}}}_{{\mathbf{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{}}{{{\mathbf{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{-}}{\mathbf{2}}{\mathbf{g}}{\mathbf{\u2206}}{\mathbf{y}}}$

**(a)**

In the absence of the air drag, the drop is falling freely.

Let the displacement downwards be in the negative y-axis.

Calculate the speed a spherical rain drop would achieve falling from 5.00 km (a) in the absence of air drag (b) with air drag. Take the diameter of the drop to be 4 mm, the density to be 1.00x10^{3} kg/m^{3}, and the surface area to be πr^{2}.

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