$A$ body falling freely under gravity passes two points $30 \ m$ apart in $1 \ s$. From what point above the upper point did it begin to fall? (Take $g = 10 \ m \ s^{-2}$)

$A$ ball is dropped on the floor from a height of $10 \, m$. It rebounds to a height of $2.5 \, m$. If the ball is in contact with the floor for $0.01 \, s$,the average acceleration during contact is:

Write the equation for the distance covered by an object falling freely during the $n^{th}$ second.

Assertion: The zero velocity of a particle at any instant always implies zero acceleration at that instant.
Reason: $A$ body is momentarily at rest when it reverses its direction of motion.
The correct option among the following is:

$A$ person sitting on the ground floor of a building notices through a window of height $1.5 \; m$ that a ball dropped from the roof of the building crosses the window in $0.1 \; s$. What is the velocity (in $m/s$) of the ball when it is at the topmost point of the window? (Take $g = 10 \; m/s^2$)

$A$ 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 $h$)