$A$ stone thrown down with a speed $u$ takes a time $t_1$ to reach the ground,while another stone thrown upwards from the same point with the same speed takes time $t_2$. The maximum height the second stone reaches from the ground is

$A$ stone is dropped from a height $h$. Simultaneously,another stone is thrown up from the ground which reaches a height $4h$. The two stones cross each other after time:

$A$ man is standing on top of a building $100\, m$ high. He throws two balls vertically,one at $t = 0$ and the other after a time interval $\Delta t$ (less than $2\, s$). The second ball is thrown at a velocity half that of the first. The vertical gap between the first and second ball is $15\, m$ at $t = 2\, s$. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throws.

$A$ ball is thrown vertically upwards with a velocity $u$. What is the distance covered by it in the last $t$ seconds of its upward motion?

$Assertion$: 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.

$A$ parachutist after bailing out falls $50\, m$ without friction. When the parachute opens,it decelerates at $2\, m/s^2$. He reaches the ground with a speed of $3\, m/s$. At what height did he bail out (in $, m$)?