$A$ body is projected vertically up with a speed $v_0$ under gravity. The average speed of the body over a time $t = \frac{v_0}{g}$ calculated from the instant of projection is:

$A$ stone thrown upward with a speed $u$ from the top of a tower reaches the ground with a velocity $3u$. The height of the tower is (in $u^2/g$)

$A$ particle is dropped under gravity from rest from a height $h$ $(g = 9.8 \ m/s^2)$. If it travels a distance of $9h/25$ in the last second of its motion,the height $h$ is ......... $m$.

$A$ stone is dropped from the top of a tower of height $h = 60 \ m$. Simultaneously,another stone is projected vertically upwards from the foot of the tower. They meet at a height $\frac{2h}{3}$ from the ground level. Find the initial velocity of the stone projected upwards. (Take $g = 10 \ ms^{-2}$) (in $ms^{-1}$)

$A$ player throws a ball upwards with an initial speed of $29.4 \; m s^{-1}$. To what height does the ball rise and after how long (in seconds) does the ball return to the player's hands? (Take $g = 9.8 \; m s^{-2}$ and neglect air resistance).

$A$ body is released from the top of a tower of height $h$. It takes $t$ seconds to reach the ground. Where will the body be after time $t/2$ seconds?