$A$ 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 $g = 10 \; m s^{-2}$.

$A$ ball falls vertically downward and bounces off a horizontal floor. The speed of the ball just before reaching the floor $(u_1)$ is equal to the speed just after leaving contact with the floor $(u_2)$,where $u_1 = u_2$. The corresponding magnitudes of accelerations are denoted respectively by $a_1$ and $a_2$. The air resistance during motion is proportional to speed and is not negligible. If $g$ is the acceleration due to gravity,then:

$Assertion$ : Two balls of different masses are thrown vertically upward with the same speed. They will pass through their point of projection in the downward direction with the same speed.
$Reason$ : The maximum height and downward velocity attained at the point of projection are independent of the mass of the ball.

$A$ 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).

$A$ player throws a ball upwards with an initial speed of $29.4\; m s^{-1}$. Choose the $x=0\; m$ and $t=0\; s$ to be the location and time of the ball at its highest point. Let the vertically downward direction be the positive direction of the $x$-axis. Determine the signs of position,velocity,and acceleration of the ball during its upward and downward motion.

$A$ parachutist,after bailing out,falls $20\,m$ without friction. When the parachute opens,it decelerates at $2\,m/s^2$. He reaches the ground with a speed of $4\,m/s$. At what height did he bail out (in $,m$)?