## University Physics with Modern Physics (14th Edition)

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...