\(\frac{9.8 \times 2}{9.8}={t}^{2}\)

\({t}=\sqrt{2}\, {sec}\)

\(\Delta\) \(t:\) time interval between drops

\({h}=\frac{1}{2} {g}(\sqrt{2}-\Delta {t})^{2}\)

\(0=\frac{1}{2} {g}(\sqrt{2}-2 \Delta {t})^{2}\)

\(\Delta {t}=\frac{1}{\sqrt{2}}\)

\({h}=\frac{1}{2} {g}\left(\sqrt{2}-\frac{1}{\sqrt{2}}\right)^{2}=\frac{1}{2} \times 9.8 \times \frac{1}{2}=\frac{9.8}{4}=2.45\, {m}\)

\({H}-{h}=9.8-2.45\)

\(=7.35\, {m}\)