$A$ ball is thrown vertically upwards with a speed of $10 \ m/s$ from the top of a tower $200 \ m$ high and another is thrown vertically downwards with the same speed simultaneously. The time difference between them in reaching the ground (in $s$) is: $(g = 10 \ m/s^2)$

$A$ body is thrown vertically upwards. If air resistance is to be taken into account,then the time during which the body rises is

$A$ body is projected vertically upward from the surface of the Earth. If the upward direction is taken as positive,then the acceleration of the body during its upward and downward journey are respectively:

$A$ ball is dropped downwards. After $1$ second,another ball is dropped downwards from the same point. What is the distance between them after $3$ seconds?

From $18 \text{ m}$ height above the ground,a ball is dropped from rest. The height above the ground at which the magnitude of velocity equals the magnitude of acceleration (in the same set of units) due to gravity is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$ and neglect the air resistance)

$A$ boy is standing on top of a tower of height $85 \, m$ and throws a ball in the vertically upward direction with a certain speed. If $5.25 \, s$ later he hears the ball hitting the ground,then the speed with which the boy threw the ball is .......... $m/s$ (take $g = 10 \, m/s^2$ and speed of sound in air $= 340 \, m/s$).