The vertical height of point $P$ above the ground is twice that of $Q$. $A$ particle is projected downward with a speed of $5 \, m/s$ from $P$ and at the same time another particle is projected upward with the same speed from $Q$. Both particles reach the ground simultaneously,then:

$A$ 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.

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:

$Assertion$ : $A$ body can have acceleration even if its velocity is zero at a given instant of time.
$Reason$ : $A$ body is momentarily at rest when it reverses its direction.

Time taken by an object falling from rest to cover the height of $h_1$ and $h_2$ is respectively $t_1$ and $t_2$,then the ratio of $t_1$ to $t_2$ is:

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)