$A$ ball is projected from a point $O$ as shown in the figure. It will strike the ground after ........ $s$ $\left(g=10 \, m/s^2\right)$.

$A$ cricket bowler releases the ball in two different ways:
$(a)$ giving it only horizontal velocity,and
$(b)$ giving it horizontal velocity and a small downward velocity.
The speed $V_s$ at the time of release is the same. Both are released at a height $H$ from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

$A$ bowling machine placed at a height $h$ above the earth's surface releases different balls with different angles but with the same velocity $10 \sqrt{3} \text{ m s}^{-1}$. All these balls' landing velocities make angles of $30^{\circ}$ or more with the horizontal. Find the height $h$ (in meters) (acceleration due to gravity $g = 10 \text{ m s}^{-2}$).

$A$ body is projected horizontally from the top of a tower of height $180 \ m$ with a velocity of $20 \ ms^{-1}$. If acceleration due to gravity is $10 \ ms^{-2}$,then match the following:

List-$I$ (Kinematic Variable)	List-$II$ (Value)
$A$. Velocity of the body after $1 \ s$ (in $ms^{-1}$)	$I$. $5$
$B$. Horizontal displacement of the body after $1 \ s$ (in $m$)	$II$. $20$
$C$. Vertical displacement of the body after $1 \ s$ (in $m$)	$III$. $10$
$D$. Vertical velocity of the body after $1 \ s$ (in $ms^{-1}$)	$IV$. $22.4$

The correct answer is

From the top of a tower of height $40 \ m,$ a ball is projected upwards with a speed of $20 \ m/s$ at an angle $30^{\circ}$ to the horizontal. The ball will hit the ground in time ........ $s$ (Take $g = 10 \ m/s^2$)

$A$ cannon placed on a cliff at a height of $375 \ m$ fires a cannonball with a velocity of $100 \ ms^{-1}$ at an angle of $30^{\circ}$ above the horizontal. The horizontal distance between the cannon and the target is (Acceleration due to gravity $= 10 \ ms^{-2}$)