\( \Rightarrow \;\;t = \sqrt {\frac{{2 \times 5}}{{10}}} \)\( = 1\;\sec \)

As the water drops fall at regular intervals from a tap therefore time difference between any two drops \( = \frac{1}{2}\;\sec \)

In this given time, distance of second drop from the tap

\( = \frac{1}{2}g{\left( {\frac{1}{2}} \right)^2} = \frac{5}{4} = 1.25\,m\)

Its distance from the ground \( = 5 - 1.25 = 3.75\;m\)