$A$ 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 $g = 10 \, m/s^2$)

Water drops fall at regular intervals from a tap which is $5\,m$ above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?

$A$ body is projected vertically upward from the surface of the Earth. If the upward direction is taken as positive,then the acceleration of the body during its upward and downward journey are respectively:

$A$ body is thrown vertically upwards and reaches its maximum height in $t$ seconds. The total time from the time of projection to reach a point at half of its maximum height while returning (in seconds) 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.

Due to the presence of air resistance, if a body dropped from a height of $20 \ m$ reaches the ground with a speed of $18 \ m/s$, then the time taken by the body to reach the ground is nearly (in $s$)