The maximum horizontal range of a projectile is $160 \ m$. When the projectile is thrown with the same speed at an elevation of $30^{\circ}$ from the horizontal,it will reach a maximum height of ......... $m$.

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

$A$ particle is projected with velocity $2 \sqrt{gh}$ at an angle $60^{\circ}$ to the horizontal so that it just clears two walls of equal height $h$ which are at a distance $2h$ from each other. The time taken by the particle to travel between these two walls is

$A$ projectile is given an initial velocity of $(\hat{i}+2 \hat{j}) \text{ ms}^{-1}$. The equation of its path is $(g=10 \text{ ms}^{-2})$

$A$ body is projected at an angle of $60^{\circ}$ with the horizontal such that the vertical component of its initial velocity is $40 \ m \ s^{-1}$. The magnitude of velocity of the projectile at one quarter of its time of flight is nearly (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m \ s^{-1}$)

$A$ cricket ball is thrown at a speed of $28 \; m/s$ in a direction $30^{\circ}$ above the horizontal. Calculate:
$(a)$ the maximum height,
$(b)$ the time taken by the ball to return to the same level,and
$(c)$ the distance from the thrower to the point where the ball returns to the same level.