What is the distance covered by a freely falling body in the first second (in $m$)? $(g = 10 \ m/s^2)$

$A$ body is thrown vertically upwards. If air resistance is to be taken into account,then the time during which the body rises is

$A$ particle is projected from the ground in a vertically upward direction at $t = 0$,with an initial velocity of $48\, m/s$. The distance travelled by the particle in the $5^{th}$ second is......$m$ $(g = 10\, m/s^2)$.

$A$ ball is projected vertically upwards from the ground. It reaches a height $h$ in time $t_1$,continues its motion,and then takes a time $t_2$ to reach the ground. The height $h$ in terms of $g, t_1$,and $t_2$ is ($g =$ acceleration due to gravity).

When a ball is thrown up vertically with velocity $u,$ it attains a maximum height $H.$ What should be the velocity so that maximum height becomes $2H$?

Two bodies $A$ (of mass $1\,kg$) and $B$ (of mass $3\,kg$) are dropped from heights of $16\,m$ and $25\,m$ respectively. The ratio of the time taken by them to reach the ground is: