In this problem, we need to consider the terminal velocity of the droplets, which is reached when the gravitational force is balanced by the drag force acting on the droplet. Terminal velocity is related to the square of the droplet's radius.

The relationship between the terminal velocity (v) and the radius (r) of the droplet is given by:

$$ v \propto r^2 $$

Initially, there are 8 equal drops of water, each with radius r and velocity 10 cm/s. When these droplets coalesce, they form a single droplet with a larger radius R. The volume of the new droplet should be equal to the total volume of the 8 smaller droplets.

Using the volume formula for spheres, we can write the relationship between the radii as:

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

Solving for R, we get:

$$ R = 2r $$

Now, we can use the relationship between the terminal velocities and radii of the droplets:

$$ \frac{v_1}{v_2} = \left(\frac{r}{R}\right)^2 $$

Given the initial terminal velocity of 10 cm/s for the smaller droplets ($v_1$) and the relationship between r and R:

$$ \frac{10}{v_2} = \left(\frac{1}{2}\right)^2 $$

Solving for the new terminal velocity ($v_2$):

$$ v_2 = 40 \mathrm{~cm} / \mathrm{s} $$

The new terminal velocity after the droplets coalesce is 40 cm/s.