Four projectiles are fired with the same velocities at angles $25^{\circ}, 40^{\circ}, 55^{\circ}$ and $70^{\circ}$ with the horizontal. The range of the projectile will be largest for the one projected at angle $.......$ . (in $^{\circ}$)

$A$ stone is projected vertically upwards with speed '$v$'. Another stone of same mass is projected at an angle of $60^{\circ}$ with the vertical with the same speed '$v$'. The ratio of their potential energies at the highest points of their journey is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$

It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point $P$ at a horizontal distance $r$ from the point of projection. If $t_1$ and $t_2$ are times taken to reach this point in two possible ways,then the product $t_1 t_2$ is proportional to

$A$ particle is projected with an angle of projection $\theta$ to the horizontal. The trajectory passes through the points $(P, Q)$ and $(Q, P)$ referred to horizontal and vertical axes ($x$-axis and $y$-axis respectively). The angle of projection $\theta$ is given by:

The angle of projection for a projectile to have the same horizontal range and maximum height is

Match the columns:

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