$A$ particle is projected from the ground in a vertically upward direction at $t = 0$,with an initial velocity of $48\, m/s$. The distance travelled by the particle in the $5^{th}$ second is......$m$ $(g = 10\, m/s^2)$.

$A$ 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:

Water drops fall from a tap onto the floor, $5 \ m$ below, at regular intervals of time. The first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from the ground at the instant when the first drop strikes the ground is . . . . . . $m$. $(g = 10 \ m/s^2)$

For a particle projected vertically upwards under gravity,it travels equal distance during the $5^{th}$ and $6^{th}$ second of its motion. Find its projection speed in $m/s$ $(g = 9.8\,m/s^2)$.

$A$ body is projected vertically upwards. The times corresponding to a specific height while ascending and while descending are $t_{1}$ and $t_{2}$ respectively. Then the velocity of projection is (where $g$ is acceleration due to gravity).

$A$ ball is projected upwards from a height $h$ above the surface of the earth with velocity $v$. The time at which the ball strikes the ground is