The ratio of the displacements of a freely falling body during the second and fifth seconds of its motion is

The time of flight of a vertically projected stone is $8 \,s$. The position of the stone after $6 \,s$ from the ground is (Acceleration due to gravity $= 10 \,m/s^2$). (in $\,m$)

Assertion $(A)$: The displacement of a vertically projected body during the last second of its upward motion is $\frac{g}{2}$.
Reason $(R)$: For a vertically projected body,the acceleration decreases gradually and becomes $\frac{g}{2}$ during the last second of upward motion.

Two bodies begin a free fall from the same height at a time interval of $N \, s$. If the vertical separation between the two bodies is $1 \, m$ after $n \, s$ from the start of the first body,then $n$ is equal to:

Water drops fall from a tap onto the floor,$5 \ m$ below,at regular intervals of time. The first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from the ground at the instant when the first drop strikes the ground is . . . . . . $m$. $(g = 10 \ m/s^2)$

Water drops fall at regular intervals from a tap which is $5\,m$ above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?