$A$ stone is thrown vertically upward with an initial velocity $v_0$. The distance travelled in time $t = \frac{1.5 v_0}{g}$ is

$A$ ball is dropped from a height $h$ above the ground. Neglecting air resistance,its velocity $(v)$ varies with its height $(y)$ above the ground as:

$A$ stone falls freely under gravity. It covers distances $h_1, h_2$ and $h_3$ in the first $5 \ s$,the next $5 \ s$ and the next $5 \ s$ respectively. The relation between $h_1, h_2$ and $h_3$ is:

$A$ body is projected vertically upward with a speed of $40 \, m/s$. The distance travelled by the body in the last second of its upward journey is ........... $m$. [Take $g = 9.8 \, m/s^2$ and neglect the effect of air resistance]

Two stones begin to fall from rest from the same height,with the second stone starting to fall $\Delta t$ seconds after the first. The distance of separation between the two stones becomes $H$,$t_0$ seconds after the first stone starts its motion. Then $t_0$ is equal to:

$A$ body falling freely under gravity passes two points $30 \ m$ apart in $1 \ s$. From what point above the upper point did it begin to fall? (Take $g = 10 \ m \ s^{-2}$)