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 thrown vertically upwards. It passes the same height at $t_1 = 2 \; s$ and $t_2 = 10 \; s$. The height is:

$A$ body is projected vertically upwards with a velocity $u$. It passes through a certain point above the ground after $t_1$ seconds. The time interval after which the body passes through the same point during the return journey is

$A$ body,thrown upwards with some velocity,reaches a maximum height of $50 \,m$. Another body with double the mass,thrown up with double the initial velocity,will reach a maximum height of............$m$.

$A$ particle is released from rest from a tower of height $3h$. The ratio of times taken to fall through successive equal heights $h$,i.e.,$t_1 : t_2 : t_3$ is

Assertion $(A)$: The displacement of a vertically projected body during the last second of its upward motion is $\frac{g}{2}$.
Reason $(R)$: For a vertically projected body,the acceleration decreases gradually and becomes $\frac{g}{2}$ during the last second of upward motion.