$A$ ball is thrown upwards. It takes $4 \, s$ to reach back to the ground. Its initial velocity is........$m/s$.

With what speed should a body be thrown upwards so that the distances traversed in the $5^{\text{th}}$ second and $6^{\text{th}}$ second are equal? (Take $g = 9.8 \, m/s^2$)

$A$ ball is thrown vertically upwards with a velocity of $19.6 \, m/s$ from the top of a tower. The ball strikes the ground after $6 \, s$. The height from the ground up to which the ball can rise is given by $\left(\frac{k}{5}\right) \, m$. The value of $k$ is ..... (use $g = 9.8 \, m/s^2$)

The initial and final velocities of a body projected vertically from the ground are $20 \,ms^{-1}$ and $18 \,ms^{-1}$ respectively. The maximum height reached by the body is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,m$)

$A$ ball (initially at rest) is released from the top of a tower. The ratio of work done by the force of gravity, in the first, second, and third seconds is

The distance covered by a stone dropped from the top of a building in the last second of its motion is $0.36$ times the total distance travelled by it. The height of the building is (acceleration due to gravity $g = 9.8 \ m/s^2$). (in $m$)