Using the following data, estimate the speed of vertically falling...

The terminal velocity $v_t$ of a spherical drop falling through a viscous medium is given by Stokes' law:

$v_t = \dfrac{2}{9} \dfrac{r^2 (\rho - \sigma) g}{\eta}$

Given:
Radius of the raindrop, $r = 0.02 \text{ cm} = 2 \times 10^{-4} \text{ m}$
Density of water, $\rho = 1000 \text{ kg/m}^3$
Density of air, $\sigma \approx 0$ (negligible compared to water)
Viscosity of air, $\eta = 1.8 \times 10^{-4} \text{ poise} = 1.8 \times 10^{-5} \text{ Pa}\cdot\text{s}$ (since $1 \text{ poise} = 0.1 \text{ Pa}\cdot\text{s}$)
Acceleration due to gravity, $g = 10 \text{ m/s}^2$

Substituting the values into the formula:

$v_t = \dfrac{2}{9} \times \dfrac{(2 \times 10^{-4})^2 \times 1000 \times 10}{1.8 \times 10^{-5}}$

$v_t = \dfrac{2}{9} \times \dfrac{4 \times 10^{-8} \times 10^4}{1.8 \times 10^{-5}}$

$v_t = \dfrac{8 \times 10^{-4}}{16.2 \times 10^{-5}}$

$v_t = \dfrac{80}{16.2} \approx 4.94 \text{ m/s}$

Rounding to the nearest integer, the speed of the vertically falling raindrops is approximately $5 \text{ m/s}$.