The maximum horizontal range of a projectile is $400\;m$. What is the maximum height attained by it (in $;m$)?

$A$ body is projected with a velocity $u$ at an angle $\theta$ with the horizontal. If the time of ascent of the body is $1 \ s$,then the maximum height it can reach is (Take $g = 10 \ m/s^2$) (in $m$)

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 thrown with a velocity of $10 \sqrt{2} \ m/s$ at an angle of $45^{\circ}$ with the horizontal. The time interval between the moments when the speed is $\sqrt{125} \ m/s$ is $(g = 10 \ m/s^2)$. (in $s$)

$A$ body is thrown horizontally from the top of a tower of height $5 \, m$. It touches the ground at a distance of $10 \, m$ from the foot of the tower. The initial velocity of the body is ......... $m/s$ $(g = 10 \, m/s^2)$.

The angle of projection of a projectile for which its initial kinetic energy becomes half at its maximum height is (in $^{\circ}$)