$A$ 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 released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies,two seconds after the release of the second body is.......$m$

An iron ball and a wooden ball of the same radius are released from a height $h$ in vacuum. The time taken by both of them to reach the ground is equal based on:

From a building,two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ is thrown downwards with the same speed $u$ (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground,then:

$A$ ball falls vertically downward and bounces off a horizontal floor. The speed of the ball just before reaching the floor $(u_1)$ is equal to the speed just after leaving contact with the floor $(u_2)$,where $u_1 = u_2$. The corresponding magnitudes of accelerations are denoted respectively by $a_1$ and $a_2$. The air resistance during motion is proportional to speed and is not negligible. If $g$ is the acceleration due to gravity,then:

Water drops fall at regular intervals from a tap which is $5\,m$ above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?