A stone is dropped into a well of $45m$ depth. The sound of splash due to the stone striking the water surface in the wall is heard after $3.125s$. From the given data, the velocity of sound in air is $(g=10m/s^2)$

The correct option is C 360 $m/s$
Time taken by stone to reach the bottom of the well is given by using $s=ut+\frac{1}{2}a{t}^2$
Where, $s=45m$ and $u=0$
$\therefore t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{90}{10}}=3.0s$
So time taken by sound to travel $45m$ is $(3.125-3)s=0.125s$
$\therefore$ velocity of sound in air, $v_{air}=\frac{\text{distance}}{\text{time}}$
$=\frac{45}{.125}$
$v_{air}=360m/s$