$A$ 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 velocity $u$ of each ball is ....... $m/s$.

With what velocity should a ball be projected vertically so that the distance covered by it in the $5^{th}$ second is twice the distance it covers in its $6^{th}$ second? (Take $g = 10 \, m/s^2$)

$A$ particle is projected upwards. The times corresponding to height $h$ while ascending and while descending are $t_1$ and $t_2$ respectively. The velocity of projection will be

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:

$A$ body is thrown vertically upwards and reaches its maximum height in $t$ seconds. The total time from the time of projection to reach a point at half of its maximum height while returning (in seconds) is: