$A$ $0.4\,kg$ mass takes $8\,s$ to reach the ground when dropped from a certain height $P$ above the surface of the Earth. The loss of potential energy in the last second of the fall is $..............J$. [Take $g = 10\,m/s^2$]

$A$ ball is projected vertically upward with an initial velocity of $50 \; ms^{-1}$ at $t = 0 \; s$. At $t = 2 \; s$,another ball is projected vertically upward with the same velocity. At $t = \dots \; s$,the second ball will meet the first ball $(g = 10 \; ms^{-2})$.

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

If a stone thrown vertically upwards from a bridge with an initial velocity of $5 \,m \,s^{-1}$ strikes the water below the bridge in a time of $3 \,s$, then the height of the bridge above the water surface is (Acceleration due to gravity $= 10 \,m \,s^{-2}$) (in $\,m$)

The ratio of the distances travelled by a freely falling body in the $1^{\text{st}}$,$2^{\text{nd}}$,$3^{\text{rd}}$,and $4^{\text{th}}$ second is:

$A$ ball is dropped from a height $H$ from rest. The ball travels $\frac{H}{2}$ in the last $1.0 \ s$. The total time taken by the ball to hit the ground is: (in $s$)