A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three second of its motion. The stone remains in the air for

The correct option is B $5s$
Given that the stone falls from rest, so $u=0$
Now distance covered in the last second is equal to the distance covered in the first three seconds of the motion. Thus,
By using second equation of motion, we get
$\Rightarrow d_3=0+\frac{1}{2}g{t}^2$
$=\frac{1}{2}\times 10\times 9=45$
Use $S_n=u+\frac{1}{2}a(2n-1)$, to find distance covered at any $n^{th}$ second.
$\Rightarrow S_t=u+(2t-1)\times \frac{10}{2}$
According to question,
$d_3=S_t$
$\Rightarrow 5(2t-1)=45$
$\therefore t=5s$