A stone is dropped from the tower covers a distance “7 S” in last second where S is distance covered in first second. How much time does the stone take to reach ground? (g=10m/s ² )

$\begin{array}{l}\text{ For freely falling }\mathrm{U}=0,\mathrm{a}=+\mathrm{g}\text{ let }\mathrm{n}\text{ ' is last sec}\\ dis\tan ce covered in nth second {\mathrm{S}}_{\mathrm{n}}=\mathrm{U}+\mathrm{a}\left(\mathrm{n}-\frac{1}{2}\right)\\ \text{ Then distance in first }\sec S_1=g\left(1-\frac{1}{2}\right)\Rightarrow S=\frac{g}{2}---\left(1\right)\\ \text{ Then distance in last }\sec S_2=g\left(n-\frac{1}{2}\right)\Rightarrow 7S=g\left(\frac{2n-1}{2}\right)---\left(2\right)\\ substitute equation \left(1\right) in \left(2\right)\\ 7\left(\frac{\mathrm{g}}{2}\right)=\frac{\mathrm{g}}{2}(2\mathrm{n}-1)\\ \Rightarrow 2\mathrm{n}-1=7\\ \Rightarrow \mathrm{n}=4\\ \text{ Hence the time of fall is }4\text{ sec }\end{array}$