What will be the ratio of the distance moved by a freely falling body from rest in $4^{th}$ and $5^{th}$ seconds of journey?

$A$ body falling from rest has a velocity $v$ after it falls through a height $h$. The distance it has to fall down further for its velocity to become double,will be $...............\,h$.

$A$ ball (initially at rest) is released from the top of a tower. The ratio of work done by the force of gravity, in the first, second, and third seconds is

From the top of a tower,a particle is thrown vertically downwards with a velocity of $10\; m/s$. The ratio of the distances covered by it in the $3^{rd}$ and $2^{nd}$ seconds of the motion is (Take $g = 10\; m/s^2$).

$A$ ball of mass $m$ is thrown vertically upwards. Assume the force of air resistance has magnitude proportional to the velocity,and direction opposite to the velocity. At the highest point,the ball's acceleration is

$A$ body is dropped from a tower of height $h$. It reaches the ground at $t = 4\, s$. What is the height of the tower in $m$?