$A$ bullet is fired from a cannon with a velocity of $500 \, m/s$. If the angle of projection is $15^\circ$ and $g = 10 \, m/s^2$,then the horizontal range is:

An object is projected such that it attains maximum range. Another body is projected to reach maximum height. If both objects reach the same maximum height,what is the ratio of their initial velocities?

$A$ projectile $A$ is thrown at an angle $30^{\circ}$ to the horizontal from point $P$ with velocity $v_1$. At the same time,another projectile $B$ is thrown with velocity $v_2$ upwards from point $Q$,which is vertically below the highest point that $A$ would reach. For $B$ to collide with $A$,the ratio $\frac{v_2}{v_1}$ should be:

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

Match the columns:

Column-$I$ $(R/H_{max})$	Column-$II$ (Angle of projection $\theta$)
$A. 1$	$1. 60^o$
$B. 4$	$2. 30^o$
$C. 4\sqrt{3}$	$3. 45^o$
$D. 4/\sqrt{3}$	$4. \tan^{-1}(4) = 76^o$

The equation of a projectile is $y = \sqrt{3}x - \frac{x^2}{2}$. The velocity of projection is: