1. Use Stokes' law for terminal velocity

For a small spherical droplet falling through air, terminal velocity is given by

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

where $r$ is the radius of the droplet, and the other quantities are constants for the same medium.

So,

$v_t \propto r^2$

2. Find the radius of the new droplet

When 8 identical droplets coalesce, volume is conserved.

If original radius is $r$, then

$8 \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$

where $R$ is the radius of the new drop.

Thus,

$R^3 = 8r^3$

$R = 2r$

3. Relate the new terminal velocity to the old one

Since terminal velocity varies as $r^2$,

$\frac{V}{v} = \left(\frac{R}{r}\right)^2 = 2^2 = 4$

Given

$v = 10\ \text{cm/s}$

therefore

$V = 4 \times 10 = 40\ \text{cm/s}$

4. Final answer

The new terminal velocity is

$\boxed{40\ \text{cm/s}}$