Two bodies of different masses $m_a$ and $m_b$ are dropped from two different heights $a$ and $b$. The ratio of the time taken by the two to cover these distances is

$A$ body is released from the top of a tower of height $h$. It takes $t$ seconds to reach the ground. Where will the body be after time $t/2$ seconds?

$A$ ball of mass $0.5 \; kg$ is dropped from a height of $10 \; m$. The height from the ground,at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity,is $\dots \; m$. (Use $g = 10 \; m/s^2$).

Time taken by an object falling from rest to cover the height of $h_1$ and $h_2$ is respectively $t_1$ and $t_2$,then the ratio of $t_1$ to $t_2$ is:

From $18 \text{ m}$ height above the ground, a ball is dropped from rest. The height above the ground at which the magnitude of velocity equals the magnitude of acceleration (in the same set of units) due to gravity is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$ and neglect the air resistance)

From a balloon moving upwards with a velocity of $12\,ms^{-1}$,a packet is released when it is at a height of $65\,m$ from the ground. The time taken by it to reach the ground is...........$s$ $(g = 10\,ms^{-2})$.