$A$ body $A$ is projected upwards with a velocity of $98 \, m/s$. $A$ second body $B$ is projected upwards with the same initial velocity but after $4 \, s$. Both the bodies will meet after how many seconds (from the start of body $A$)?

$A$ body is falling freely from the top of a tower of height $125 \ m$. The distance covered by the body during the last second of its motion is $x \%$ of the height of the tower. Then $x$ is (Acceleration due to gravity $= 10 \ m/s^2$)

$A$ ball is released from a height $h$. If $t_{1}$ and $t_{2}$ are the times required to complete the first half and the second half of the distance respectively,then choose the correct relation between $t_{1}$ and $t_{2}$.

From $18 \text{ m}$ height above the ground,a ball is dropped from rest. The height above the ground at which the magnitude of velocity equals the magnitude of acceleration (in the same set of units) due to gravity is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$ and neglect the air resistance)

$A$ body thrown vertically up with initial velocity $52 \,m/s$ from the ground passes twice a point at $h$ height above at an interval of $10 \,s$. The height $h$ is .........$m$ $(g=10 \,m/s^2)$.

$A$ body falling from a high minaret travels $40 \ m$ in the last $2 \ s$ of its fall to the ground. The height of the minaret in meters is (take $g = 10 \ m/s^2$):