$A$ body is dropped from a certain height $h$ ($h$ is very large) and a second body is thrown downward with a velocity of $5 \, m/s$ simultaneously. What will be the difference in the heights of the two bodies after $3 \, s$?

$A$ 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:

$A$ ball is projected vertically up from the ground. Boy $A$, standing at the window of the first floor of a nearby building, observes that the time interval between the ball crossing him while going up and the ball crossing him while going down is $2 \ s$. Another boy $B$, standing on the second floor, notices the time interval between the ball passing him twice, during up and down motion, is $1 \ s$. Calculate the difference between the vertical positions of boy $B$ and boy $A$. (Assume $g = 10 \ m \ s^{-2}$) (in $m$)

Write the equation of velocity of an object falling freely from height $h$.

$A$ ball is thrown upward with speed $10 \, m/s$ from the top of a tower and reaches the ground with a speed $20 \, m/s$. The height of the tower is ......... $m$. (Take $g = 10 \, m/s^2$)

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.