$A$ ball is released from the top of a tower of height $h$ meters. It takes $T$ seconds to reach the ground. What is the position of the ball at $T/3$ seconds?

$A$ body is projected vertically upwards. The times corresponding to a specific height while ascending and while descending are $t_{1}$ and $t_{2}$ respectively. Then the velocity of projection is (where $g$ is acceleration due to gravity).

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:

Two balls $A$ and $B$ are thrown with the same velocity $u$ from the top of a tower. Ball $A$ is thrown vertically upwards and ball $B$ is thrown vertically downwards. If a ball $C$ is thrown with the same velocity $u$ in the horizontal direction from the top of the tower,and if $t_A = 6 \ s$ and $t_B = 2 \ s$,then the time $t_C$ taken by ball $C$ to reach the ground is equal to........$s$.

$A$ body is projected vertically upwards with a velocity $u$ from the top of a tower. The time taken by it to reach the ground is $n$ times the time taken by it to reach the highest point in its path. The height of the tower is:

$A$ stone is dropped into water from a bridge $44.1 \ m$ above the water. Another stone is thrown vertically downward $1 \ s$ later. Both strike the water simultaneously. What was the initial speed of the second stone in $m/s$?