Water droplets are coming from an open tap at a particular rate....

In ${4}^{th}sec$, ${1}^{st}$ drop will travel

$⇒ \dfrac{1}{2} \times \left(9 . 8\right) \times {\left(4\right)}^{2} = 78 . 4m$

$∴ {2}^{nd}$ drop would have travelled

$⇒ 78 . 4 - 34 . 3 = 44 . 1m$

Time for ${2}^{nd}$ drop

$\dfrac{1}{2}\left(9 . 8\right){t}^{2} = 44 . 1$

$t = 3sec$

$∴$ Each drop have time gap of $1sec$.

$∴ 1$ drop per second.