$A$ very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is $5\, m$,the number of balls thrown per minute is (take $g = 10\, ms^{-2}$):-

$A$ ball is thrown upward with an initial velocity $V_0$ from the surface of the earth. The motion of the ball is affected by a drag force equal to $m\gamma v^2$ (where $m$ is the mass of the ball,$v$ is its instantaneous velocity,and $\gamma$ is a constant). The time taken by the ball to rise to its zenith is:

$Assertion$: If a body is thrown upwards,the distance covered by it in the last second of upward motion is about $5 \ m$ irrespective of its initial speed.
$Reason$: The distance covered in the last second of upward motion is equal to that covered in the first second of downward motion when the particle is dropped.

$A$ body falls from a height $h = 200\,m$ (at New Delhi). The ratio of distance travelled in each $2\,s$ interval during $t = 0$ to $t = 6\,s$ of the journey is

If the distance travelled by a freely falling body in the last but one second of its motion is $5 \ m$,then the total time of motion is (Acceleration due to gravity $g = 10 \ m \ s^{-2}$) (in $s$)

$A$ stone falls from a balloon that is descending at a uniform rate of $12 \, m/s$. The displacement of the stone from the point of release after $10 \, s$ is.......$m$ $(g = 9.8 \, m/s^2)$