A small drop of water falls from rest through a large height h in air; the final velocity is

The correct option is: Almost independent of h

Let's understand why this is the case by breaking down the physics of a falling water droplet:

1. Initial Fall from Rest:
When a small water droplet starts falling from rest, it is initially accelerated downwards under the influence of gravity ($g$). At this very early stage, its velocity is small, and air resistance (drag force) is negligible.

2. The Effect of Air Resistance:
As the droplet falls and its velocity increases, it experiences an opposing drag force from the surrounding air. According to Stokes' Law, for a small spherical droplet, the viscous drag force ($F_d$) is directly proportional to its velocity ($v$):
$F_d=6\pi \eta rv$
where $\eta$ is the viscosity of air and $r$ is the radius of the droplet.

3. Reaching Terminal Velocity:
As the speed increases, the downward gravitational force ($F_g=mg$) and the upward drag force (along with any small buoyant force) eventually balance each other out. Once these forces are balanced:
Net Force = 0 ⇒ Acceleration = 0
At this point, the droplet ceases to accelerate and continues to fall at a constant velocity, known as the terminal velocity ($v_t$).

4. Dependency on the Height (h):
Because the drop falls through a "large height $h$," it reaches this terminal velocity very early in its path. For the vast majority of its long descent, it travels at this constant terminal speed. Thus, the final velocity with which it reaches the ground is equal to the terminal velocity, which is determined solely by the properties of the droplet and the air (radius, density, viscosity, gravity), and is almost independent of the height h from which it was dropped.