The equation of a projectile is given by $y = Px - Qx^2$,where $P$ and $Q$ are constants. The ratio of the maximum height to the range of the projectile is

$A$ batsman hits a ball with a velocity '$v$',making an angle of $60^{\circ}$ with the vertical. After some time, the direction of velocity makes an angle of $60^{\circ}$ with the horizontal. The speed of the ball at this instant is $[\cos(60^{\circ}) = \frac{1}{2}, \cos(30^{\circ}) = \frac{\sqrt{3}}{2}]$.

$A$ particle is projected with a velocity $v$ such that its horizontal range is twice the greatest height attained. The horizontal range is

$A$ body is projected horizontally from the top of a tower of height $180 \,m$ with a velocity of $20 \,ms^{-1}$. If acceleration due to gravity is $10 \,ms^{-2}$, then match the following:

$A$. Velocity of the body after $1 \,s$ (in $ms^{-1}$)	$I$. $5$
$B$. Horizontal displacement of the body after $1 \,s$ (in $m$)	$II$. $20$
$C$. Vertical displacement of the body after $1 \,s$ (in $m$)	$III$. $10$
$D$. Vertical velocity of the body after $1 \,s$ (in $ms^{-1}$)	$IV$. $22.4$

$A$ body is projected with a speed $u$ at an angle $\theta$ with the horizontal. The radius of curvature of the trajectory,when it makes an angle $\left(\frac{\theta}{2}\right)$ with the horizontal is ($g$ = acceleration due to gravity).

In a spring gun having a spring constant $k = 100\, \text{N/m}$,a small ball $B$ of mass $m = 100\, \text{g}$ is placed in its barrel (as shown in the figure) by compressing the spring by $x = 0.05\, \text{m}$. $A$ box is placed at a distance $d$ on the ground so that the ball falls into it. If the ball leaves the gun horizontally at a height of $h = 2\, \text{m}$ above the ground,find the value of $d$ in meters. (Take $g = 10\, \text{m/s}^2$)