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Question:

"Consider a small (less massive) raindrop, and a large (more massive) raindrop (consider both to be spheres), falling with air resistance (drag). Which will have a larger terminal speed, or will they be the same? Explain."

The answer says that the larger raindrop will have a larger terminal velocity because the velocity is given by the square root of the radius.

I'm wondering how this equation for terminal velocity was found. I know that at terminal velocity, we can say that the force of gravity equals the drag force because there is no more acceleration. But using the equation of drag force for small spheres, $F=6πaηv$, and setting this equal to $mg$, I get a terminal velocity that's proportional to mass and inversely proportional to radius. I don't understand how the mass has no effect.

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The mass do has an effect but it's implicit, though. That is, it is expressed in terms of the density of water and radius of the drop.

Also, you are right that you can find terminal velocity but equating the weight and drag force for the drop. This is given by

$$F_{\text{air}} = \frac{1}{2} \rho_\text{air} A C v^2\, ,$$

where $\rho$ is the density of air, $A$ is the cross-sectional area of the drop, $C$ is the drag coefficient, and $v$ is the speed.

The equation to solve is

$$\frac{1}{2} \rho_\text{air} \left(\pi R^2\right) C v^2 = \rho_{\text{water}} \left(\frac{4}{3}\pi R^3\right) g\, ,$$

that after solving for $v$ is

$$v = \sqrt{\frac{8}{3} \frac{\rho_\text{water} g}{\rho_\text{air} C} R} \, .$$

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