$A$ ball is thrown from the ground to clear a wall $3 \, m$ high at a distance of $6 \, m$ and falls $18 \, m$ away from the wall. The angle of projection of the ball is:

$A$ fighter jet is flying horizontally at a certain altitude with a speed of $200 \; m/s$. When it passes directly overhead an anti-aircraft gun,a bullet is fired from the gun at an angle $\theta$ with the horizontal to hit the jet. If the bullet speed is $400 \; m/s$,the value of $\theta$ will be $\dots \; ^\circ$.

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

$A$ projectile fired with initial velocity $u$ at an angle $\theta$ has a range $R$. If the initial velocity is doubled at the same angle of projection,then the new range will be:

$A$ body is thrown at an angle of $30^{\circ}$ to the horizontal with a velocity of $30\; m/s$. After $1\; s$,its velocity will be (in $m/s$) $\left(g=10\; m/s^{2}\right)$

Two particles $A$ and $B$ are projected simultaneously from a fixed point on the ground. Particle $A$ is projected on a smooth horizontal surface with speed $v$,while particle $B$ is projected in air with speed $\frac{2v}{\sqrt{3}}$. If particle $B$ hits particle $A$,the angle of projection of $B$ with the vertical is