$A$ ball is thrown vertically upwards with a velocity of $20 \; m s^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0 \; m$ from the ground. How long will it be in $seconds$ before the ball hits the ground? Take $g = 10 \; m s^{-2}$.

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

Water drops fall from a tap onto the floor,$5 \ m$ below,at regular intervals of time. The first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from the ground at the instant when the first drop strikes the ground is . . . . . . $m$. $(g = 10 \ m/s^2)$

$A$ balloon was moving upwards with a uniform velocity of $10 \, m/s$. An object of finite mass is dropped from the balloon when it was at a height of $75 \, m$ from the ground level. The height of the balloon from the ground when the object strikes the ground was around (in $m$). (Take the value of $g$ as $10 \, m/s^2$)

$A$ body dropped from the top of a tower covers a distance $7x$ in the last second of its journey,where $x$ is the distance covered in the first second. How much time does it take to reach the ground?

$A$ 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: