$\frac{9.8 \times 2}{9.8}={t}^{2}$
${t}=\sqrt{2}\, {sec}$
$\Delta$ $t:$ time interval between drops
${h}=\frac{1}{2} {g}(\sqrt{2}-\Delta {t})^{2}$
$0=\frac{1}{2} {g}(\sqrt{2}-2 \Delta {t})^{2}$
$\Delta {t}=\frac{1}{\sqrt{2}}$
${h}=\frac{1}{2} {g}\left(\sqrt{2}-\frac{1}{\sqrt{2}}\right)^{2}=\frac{1}{2} \times 9.8 \times \frac{1}{2}=\frac{9.8}{4}=2.45\, {m}$
${H}-{h}=9.8-2.45$
$=7.35\, {m}$
(જ્યાં $v$ વેગ છે)