$A$ 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 released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies,two seconds after the release of the second body is.......$m$

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)

$A$ player throws a ball upwards with an initial speed of $29.4\; m s^{-1}$. Choose the $x=0\; m$ and $t=0\; s$ to be the location and time of the ball at its highest point. Let the vertically downward direction be the positive direction of the $x$-axis. Determine the signs of position,velocity,and acceleration of the ball during its upward and downward motion.

Assertion $(A)$: The displacement of a vertically projected body during the last second of its upward motion is $\frac{g}{2}$.
Reason $(R)$: For a vertically projected body,the acceleration decreases gradually and becomes $\frac{g}{2}$ during the last second of upward motion.

$A$ body is thrown vertically upwards and takes $5 \text{ s}$ to reach maximum height. The distance travelled by the body will be same in