A stone is allowed to fall from top of a building and covers half of the height in last second of its motion. The time taken by stone to reach the bottom of building is?

The correct option is B $(2-\sqrt{2})s$
Let $‘h^{\prime }$ be the height of the tower, $‘n^{\prime }$ be the time taken to reach the ground $(h,n,>0)$.

Using the formula $S$ = $ut+\frac{1}{2}a{t}^2$

=> $h$ = $0$ + $\frac{1}{2}\left(9.8\right)n^2$…………………….(1)

According to the problem, the stone travels $\frac{h}{2}$ in the $n^{th}$ second.

So, by using $S_n$ = $u$ + $\frac{a}{2}(2n-1)$ we get

=> $\frac{h}{2}$ = $0$ + $\frac{9.8}{2}(2n-1)$ [as initial velocity is $0$]

Using (1),

$\frac{\frac{1}{2}\left(9.8\right)n^2}{2}$ = $\frac{9.8}{2}(2n-1)$

=> $\frac{n^2}{2}$ = $2n–1$

=> $n$ = $(2+\surd 2)$ s or $(2-\surd 2)$ s