The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

$A$ ball is projected with a velocity $5 \text{ m/s}$, such that its horizontal range is twice the greatest height attained. The value of the range is (in $\text{ m}$)

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 is projected at $10 \ m/s$ by making an angle of $60^{\circ}$ to the horizontal. After some time,its velocity makes an angle of $30^{\circ}$ to the horizontal. Its speed at this instant is:

What do you mean by projectile motion and projectile particle? Find the value of the position of a projectile particle at any instant of time.

The equation of motion of a projectile is $y = 12x - \frac{3}{4}x^2$. The range of the projectile is $..........\,m$.