When a ball is thrown up vertically with velocity $u,$ it attains a maximum height $H.$ What should be the velocity so that maximum height becomes $2H$?

$A$ ball is dropped from rest at time $t=0$ from a certain height. $A$ second ball is dropped from the same height at time $t=1 \,s$. At what time $t$, does the distance between the two balls become $10 \,m$ (in $s$)?

If a stone is released from a balloon rising with acceleration $a$ at the instant when its velocity is $v$,then immediately after release,the acceleration and velocity of the stone are

$A$ ball is released from the top of a tower of height $H \ m$. It takes $T \ s$ to reach the ground. The height of the ball from the ground after $\frac{T}{4} \ s$ is

$A$ man throws a ball vertically upward and it rises through $20 \; m$ and returns to his hands. What was the initial velocity $(u)$ of the ball and for how much time $(T)$ it remained in the air? $[g = 10 \; m/s^2]$

If a particle takes $t$ seconds less and acquires a velocity of $v \ ms^{-1}$ more in falling through the same distance (starting from rest) on two planets where the accelerations due to gravity are $2g$ and $8g$ respectively,then $v=$