$A$ body freely falling from rest has a velocity $v$ after it falls through a height $h$. The total distance it has to fall down from the starting point for its velocity to become double is: (in $h$)

Ball-$1$ is dropped from the top of a building from rest. At the same moment, ball-$2$ is thrown upward towards ball-$1$ with a speed $14 \,m/s$ from a point $21 \,m$ below the top of the building. How far will ball-$1$ have dropped when it passes ball-$2$? (Assume acceleration due to gravity, $g=10 \,m/s^2$.)

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 $v=$

$A$ ball is thrown vertically upwards with a velocity of $15 \ m/s$. To what maximum height will it reach? (Take $g = 10 \ m/s^2$)

$A$ stone is thrown upwards with a velocity $50 \,ms^{-1}$. Another stone is simultaneously thrown downwards from the same location with a velocity $50 \,ms^{-1}$. When the first stone is at the highest point, the velocity of the second stone is $...... \,ms^{-1}$ (Take $g = 10 \,ms^{-2}$).

$A$ body is thrown vertically upwards and takes $5 \text{ s}$ to reach maximum height. The distance travelled by the body will be same in