$A$ body at rest falls through a height $h$ with velocity $V$. If it has to fall down further for its velocity to become $3V$,the distance travelled in that interval is: (in $h$)

$A$ stone is allowed to fall from the top of a tower $100 \ m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25 \ m/s$. The two stones will meet after ......... $s$.

$A$ 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:

Galileo's law of odd numbers: "The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, $1: 3: 5: 7 \ldots$]". Prove it.

The distance covered by a stone dropped from the top of a building in the last second of its motion is $0.36$ times the total distance travelled by it. The height of the building is (acceleration due to gravity $g = 9.8 \ m/s^2$). (in $m$)

Ball-$1$ is dropped from the top of a building from rest. At the same moment, ball-$2$ is thrown upward towards ball-$1$ with a speed $14 \,m/s$ from a point $21 \,m$ below the top of the building. How far will ball-$1$ have dropped when it passes ball-$2$? (Assume acceleration due to gravity, $g=10 \,m/s^2$.)