$A$ student is throwing balls vertically upwards such that he throws the $2^{\text{nd}}$ ball when the $1^{\text{st}}$ ball reaches its maximum height. If he throws balls at an interval of $3 \ s$,what is the maximum height reached by the balls (in $m$)? (Take $g = 10 \ m/s^2$)

$A$ 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:

$A$ body at rest falls through a height $h$ with velocity $V$. If it has to fall down further for its velocity to become $3V$,the distance travelled in that interval is: (in $h$)

$A$ parachutist after bailing out falls $50\, m$ without friction. When the parachute opens,it decelerates at $2\, m/s^2$. He reaches the ground with a speed of $3\, m/s$. At what height did he bail out (in $, m$)?

$A$ ball is thrown upward from the ground with an initial speed of $v$. At the same instant, another ball is dropped from a building of height $20 \,m$. If the balls are at the same height after $0.8 \,s$, then the magnitude of $v$ is (Take, acceleration due to gravity $=10 \,ms^{-2}$) (in $\,ms^{-1}$)

If a ball is thrown vertically upwards with speed $u$,the distance covered during the last $t$ seconds of its ascent is