$A$ particle is projected vertically upwards such that it attains a height of $h$ after $5 \ s$ and $9 \ s$ of its motion. The speed of projection is $...... \ m/s$ $(g = 10 \ m/s^2)$.

$A$ ball of mass $0.5 \; kg$ is dropped from a height of $10 \; m$. The height from the ground,at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity,is $\dots \; m$. (Use $g = 10 \; m/s^2$).

$A$ body is released from the top of a tower of height $H \, m$. After $2 \, s$ it is stopped and then instantaneously released. What will be its height from the ground after the next $2 \, s$?

$A$ body is thrown vertically upwards and takes $5 \text{ s}$ to reach maximum height. The distance travelled by the body will be same in

$A$ body is dropped from a height $H$. The time taken to cover the second half of the journey is:

Two balls $A$ and $B$ are thrown with the same velocity $u$ from the top of a tower. Ball $A$ is thrown vertically upwards and ball $B$ is thrown vertically downwards. If $t_A = 6 \ s$ and $t_B = 2 \ s$,then the height of the tower is ......... $m$.