What is the distance covered by a freely falling body in the first second (in $m$)? $(g = 10 \ m/s^2)$

An iron ball and a wooden ball of the same radius are released from a height $h$ in vacuum. The time taken by both of them to reach the ground is

If a particle takes $t$ seconds less and acquires a velocity of $v \ ms^{-1}$ more in falling through the same distance (starting from rest) on two planets where the accelerations due to gravity are $2g$ and $8g$ respectively,then $v=$

$A$ body is dropped from a height $h$. If it acquires a momentum $p$ just before striking the ground,then the mass of the body is

$A$ ball falls vertically downward and bounces off a horizontal floor. The speed of the ball just before reaching the floor $(u_1)$ is equal to the speed just after leaving contact with the floor $(u_2)$,where $u_1 = u_2$. The corresponding magnitudes of accelerations are denoted respectively by $a_1$ and $a_2$. The air resistance during motion is proportional to speed and is not negligible. If $g$ is the acceleration due to gravity,then:

$A$ particle is thrown vertically upwards. It passes the same height at $t_1 = 2 \; s$ and $t_2 = 10 \; s$. The height is: