$A$ body is projected vertically up with a speed $v_0$ under gravity. The average speed of the body over a time $t = \frac{v_0}{g}$ calculated from the instant of projection is:

$A$ 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).

Water drops fall from a tap onto the floor, $5 \ m$ below, at regular intervals of time. The first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from the ground at the instant when the first drop strikes the ground is . . . . . . $m$. $(g = 10 \ m/s^2)$

$A$ body $A$ is projected upwards with a velocity of $98 \, m/s$. $A$ second body $B$ is projected upwards with the same initial velocity but after $4 \, s$. Both the bodies will meet after how many seconds (from the start of body $A$)?

$A$ balloon starts rising from the ground with an acceleration of $1.25\, m/s^2$. After $8\,s$,a stone is released from the balloon. The stone will $(g = 10\, m/s^2)$:

$A$ body is dropped from a height $h$. If it acquires a momentum $p$ just before striking the ground,then the mass of the body is