$A$ person throws balls into the air vertically upward at regular intervals of time of $1 \,s$. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is (Assume $g = 10 \,m/s^2$) (in $\,m$)

$A$ ball is dropped from the top of a $100\; m$ high tower on a planet. In the last $\frac{1}{2}\; s$ before hitting the ground,it covers a distance of $19\; m$. Acceleration due to gravity (in $m/s^2$) near the surface on that planet is:

$A$ stone is dropped from a certain height and reaches the ground in $5\, s$. If the stone is stopped after $3\, s$ of its fall and then allowed to fall again,then the time taken by the stone to reach the ground for the remaining distance is........$s$.

Two bodies begin a free fall from the same height at a time interval of $N \, s$. If the vertical separation between the two bodies is $1 \, m$ after $n \, s$ from the start of the first body,then $n$ is equal to:

$Assertion$ : Two balls of different masses are thrown vertically upward with the same speed. They will pass through their point of projection in the downward direction with the same speed.
$Reason$ : The maximum height and downward velocity attained at the point of projection are independent of the mass of the ball.

$A$ ball is thrown up vertically with a certain velocity so that it reaches a maximum height $h$. Find the ratio of the times at which it is at height $\frac{h}{3}$ while going up and coming down respectively.