terminal speed of two raindrop

Yes, you are right. We can apply this formula:

D = 1/2 C ρ A v²

where D = drag force
C = drag coefficient
ρ = density
A = cross-sectional area
v = speed

To find the terminal velocity, setting the drag force D equal to the gravity of the raindrop, i.e.
D = mg = ρ(4/3 π r³) g, where r is the radius of the raindrop
=> ρ(4/3 π r³) g = 1/2 C ρ π r² v²
=> v = √ [(8/3 r g ) / C]

Hence, v is proportional to the square root of the radius of the raindrop.
Therefore, the bigger raindrop has higher terminal speed.

Note: The drag coefficient, C, is only approximately constant for the raindrops of different sizes. The shape of the raindrops are only approximately spherical.

I do not understand your last sentence.

The force addes on the raindrop is F=mg-bv, m is the mass of the raindrop.

m=pV, V is the volume, p is density.b is the air resistance constant depend on the surface area. V=4/3(pi)r^3. (Volume of a sphere)

Asume the two raindrops are sphere in shape.

And the terminal speed, F=0. => mg=bv, b=2(pi)r^2 x c (Area of half of sphere), c is a constant.

v=mg/b=p 4/3(pi)r^3 / c [2(pi)r^2] = p/c x (2r/3)

v is proprotional to r, that mean if the r is double, the terminal speed is double.

2007-06-14 16:27:12 補充：
Since p, c are the same for the same matter.