If we can throw a ball up to a maximum height $H$,the maximum horizontal distance to which we can throw it is

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

Two bodies are projected at angles $\theta$ and $(90^\circ - \theta)$ with the same initial velocity. Find the ratio of their times of flight.

The horizontal range of a projectile projected at an angle of $45^{\circ}$ with the horizontal is $50 \ m$. The height of the projectile when its horizontal displacement is $20 \ m$ is (in $m$)

Assertion $(A)$: The range of a projectile is maximum when the angle of projection is $45^{\circ}$.
Reason $(R)$: The range of a projectile depends only on the angle of projection.

$A$ bullet is dropped from a certain height,and at the same instant,another bullet is fired horizontally from the same height. Which bullet will hit the ground first?