$A$ 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 $g = 9.8\,m/s^2$)

Simultaneously,from the top of a tower,when ball-$1$ is thrown horizontally and ball-$2$ is just dropped,in the absence of air resistance,which among the following options is correct?

The equation of motion of a projectile is $y = ax - bx^2$,where $a$ and $b$ are constants. Match the Column-$I$ with Column-$II$:

Column-$I$	Column-$II$
$i)$ The initial velocity of projection	$a)$ $\sqrt{\frac{g(1+a^2)}{2b}}$
$ii)$ The horizontal range of projectile	$b)$ $\frac{a}{b}$
$iii)$ The maximum height attained by projectile	$c)$ $\frac{a^2}{4b}$
$iv)$ The time of flight of projectile	$d)$ $a\sqrt{\frac{2}{bg}}$

In the projectile motion of a particle on a level ground,which of the following remains constant with reference to time and position?

$A$ projectile is projected with a velocity of $25 \, m/s$ at an angle $\theta$ with the horizontal. After $t$ seconds,its inclination with the horizontal becomes zero. If $R$ represents the horizontal range of the projectile,the value of $\theta$ will be: [use $g = 10 \, m/s^2$]

$A$ particle is projected at $60^\circ$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is