$A$ body falling under gravity covers two points $A$ and $B$ separated by $80 \ m$ in $2 \ s$. The distance of the upper point $A$ from the starting point is (use $g = 10 \ m/s^2$): (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, $g=10 \,m/s^2$.)

$A$ 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

Two spherical balls having equal masses with a radius of $5 \, \text{cm}$ each are thrown upwards along the same vertical direction at an interval of $3 \, \text{s}$ with the same initial velocity of $35 \, \text{m/s}$. These balls collide at a height of $\ldots \ldots \ldots \, \text{m}$. (Take $g = 10 \, \text{m/s}^2$)

$A$ ball is projected vertically upwards from the ground. It reaches a height $h$ in time $t_1$,continues its motion,and then takes a time $t_2$ to reach the ground. The height $h$ in terms of $g, t_1$,and $t_2$ is ($g =$ acceleration due to gravity).

$A$ body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies,two seconds after the release of the second body is.......$m$