$A$ ball is thrown vertically upwards with a velocity $u$. What is the distance covered by it in the last $t$ seconds of its upward motion?

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 $g = 9.8 \, m/s^2$)

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. Choose the correct statement.

$A$ body $A$ is projected upwards with a velocity of $98 \, m/s$. $A$ second body $B$ is projected upwards with the same initial velocity but after $4 \, s$. Both the bodies will meet after how many seconds (from the start of body $A$)?

$A$ ball is released from a height $h$. If $t_{1}$ and $t_{2}$ are the times required to complete the first half and the second half of the distance respectively,then choose the correct relation between $t_{1}$ and $t_{2}$.

$A$ 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......$m$ $(g = 10\, m/s^2)$.