$A$ cricketer hits a ball with a velocity $25\,m/s$ at $60^\circ$ above the horizontal. How far above the ground does it pass over a fielder $50\,m$ from the bat (in $,m$)? (Assume the ball is struck very close to the ground and $g = 9.8\,m/s^2$)

For an object projected from the ground with speed $u$,the horizontal range is two times the maximum height attained by it. The horizontal range of the object is ..........

$A$ 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:

$A$. 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 $ms^{-1}$)	$IV$. $22.4$

$A$ cricket ball is hit at $30^{\circ}$ with the horizontal with kinetic energy $K$. The kinetic energy at the highest point is

The vertical displacement ($y$ in metre) of a projectile in terms of its horizontal displacement ($x$ in metre) is given by $y = (\sqrt{3}x - 0.2x^2)$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ ms^{-2}$)

$A$ cricket ball is thrown at a speed of $28 \; m/s$ in a direction $30^{\circ}$ above the horizontal. Calculate:
$(a)$ the maximum height,
$(b)$ the time taken by the ball to return to the same level,and
$(c)$ the distance from the thrower to the point where the ball returns to the same level.