a
In 4 sec. \(1^{\text {st }}\) drop will travel \(\Rightarrow \frac{1}{2} \times(9.8) \times(4)^{2}=78.4 m\)

\(\therefore 2^{\text {nd }}\) drop would have travelled \(\Rightarrow 78.4-34.3=44.1 m\).

Time for \(2^{\text {nd }}\) drop

\(\frac{1}{2}(9.8) t ^{2}=44.1\)

\(t=3 sec\)

\(\therefore\) each drop have time gap of \(1 sec\)

\(\therefore 1\) drop per sec