#### Answer

The distance between the drops will increase as they fall.

#### Work Step by Step

Let $d_1$ be the first drop and let $d_2$ be the second drop. Let $t=0$ be the time when the first drop falls.
The speed of $d_1$ is $v_1 = gt$.
The speed of $d_2$ is $v_2 = g(t-1)$ when $t \geq 1$.
When $t \geq 1$:
$v_1 = gt > gt - g = g(t-1) = v_2$
Since $v_1$ is always greater than $v_2$, the distance between the drops will increase as they fall.
Note that we can make a similar argument for the following drops $d_3, d_4, d_5,$ etc...