The horizontal range of a projectile is $4\sqrt{3}$ times its maximum height. Its angle of projection will be ......... $^\circ$.

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	$d)$ $a\sqrt{\frac{2}{bg}}$

$A$ particle of mass $m$ is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

$A$ projectile is thrown with velocity $v$ at an angle $\theta$ with the horizontal. When the projectile is at a height equal to half of the maximum height,the vertical component of the velocity of the projectile is ...........

$A$ projectile is thrown such that it achieves a maximum range $R$. What are the coordinates of the point where its velocity is minimum?

$A$ body of mass $m$ is thrown upwards at an angle $\theta$ with the horizontal with velocity $v$. While rising up,the velocity of the mass after $t$ seconds will be: