$A$ 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?

An object of mass $5 \, kg$ is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of $10 \, N$ throughout the motion. The ratio of time of ascent to the time of descent will be equal to [Use $g = 10 \, ms^{-2}$]

Four marbles are dropped from the top of a tower one after the other with an interval of one second. The first one reaches the ground after $4$ seconds. When the first one reaches the ground,the distances between the first and second,the second and third,and the third and fourth marbles will be respectively:

$A$ body is thrown vertically upwards with a velocity of $19.6 \, ms^{-1}$. The position of the body after $4 \, s$ will be

If a stone thrown vertically upwards from a bridge with an initial velocity of $5 \,m \,s^{-1}$ strikes the water below the bridge in a time of $3 \,s$, then the height of the bridge above the water surface is (Acceleration due to gravity $= 10 \,m \,s^{-2}$) (in $\,m$)

When a ball is thrown up vertically with velocity $V_o$,it reaches a maximum height of $h$. If one wishes to triple the maximum height,then the ball should be thrown with what velocity?