$A$ man drops a ball from the roof of a tower of height $400 \ m$. At the same time,another ball is thrown upwards with a velocity of $50 \ m/s$ from the base of the tower. At what height from the base of the tower will they meet (in $m$)?

$A$ body is released from the top of a tower of height $h$. It takes $t$ seconds to reach the ground. Where will the body be after time $t/2$ seconds?

An object is projected upwards from the foot of a tower. The object crosses the top of the tower twice with an interval of $8 \ s$ and the object reaches the foot again after $16 \ s$. The height of the tower is [ $g = 10 \ m/s^2$ ]. (in $m$)

Two bodies having masses in the ratio $2:3$ fall freely under gravity from heights which are in the ratio $9:16$. The ratio of their linear momenta on touching the ground is

$A$ player throws a ball upwards with an initial speed of $29.4 \; m s^{-1}$. To what height does the ball rise and after how long (in seconds) does the ball return to the player's hands? (Take $g = 9.8 \; m s^{-2}$ and neglect air resistance).

$Assertion$: 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.