The time of flight of a vertically projected stone is $8 \,s$. The position of the stone after $6 \,s$ from the ground is (Acceleration due to gravity $= 10 \,m/s^2$). (in $\,m$)

$A$ stone is thrown with an initial speed of $4.9 \, m/s$ from a bridge in a vertically upward direction. It falls down into the water after $2 \, s$. The height of the bridge is .......... $m$.

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)

$A$ 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.

$A$ body is released from the top of a tower of height $h$. It takes $t$ seconds to reach the ground. Where will the body be after time $t/2$ seconds?

$A$ ball is dropped from the top of a $100\; m$ high tower on a planet. In the last $\frac{1}{2}\; s$ before hitting the ground,it covers a distance of $19\; m$. Acceleration due to gravity (in $m/s^2$) near the surface on that planet is: