$A$ particle starts from the origin at $t=0$ with an initial velocity of $3.0 \hat{i} \; m/s$ and moves in the $x-y$ plane with a constant acceleration $(6.0 \hat{i} + 4.0 \hat{j}) \; m/s^2$. The $x$-coordinate of the particle at the instant when its $y$-coordinate is $32 \; m$ is $D$ meters. The value of $D$ is

Column-$I$ (Angle of projection)	Column-$II$
$A. \theta = 45^{\circ}$	$1. \frac{K_h}{K_i} = \frac{1}{4}$
$B. \theta = 60^{\circ}$	$2. \frac{gT^2}{R} = 8$
$C. \theta = 30^{\circ}$	$3. \frac{R}{H} = 4\sqrt{3}$
$D. \theta = \tan^{-1}(4)$	$4. \frac{R}{H} = 4$

$K_i:$ initial kinetic energy,$K_h:$ kinetic energy at the highest point.

At $t = 0$,a projectile is fired from a point $O$ (taken as origin) on the ground with a speed of $50 \, m/s$ at an angle of $53^{\circ}$ with the horizontal. It passes through two points $A$ and $B$,each at a height of $75 \, m$ above the horizontal,as shown in the figure. Find the distance (in meters) of the particle from the origin at $t = 2 \, s$.

$A$ person walks in such a way that he covers equal distance in each step. The person takes $2$ steps forward towards East, then takes a right turn and walks $4$ steps towards South, then takes a right turn and walks $6$ steps towards West, and then takes a right turn and walks further. The direction of his final position after a total of $20$ steps walk with respect to his initial position is:

Two stones are projected with the same speed but making different angles with the horizontal. Their ranges are equal. If the angle of projection of one is $\pi / 3$ and its maximum height is $h_1$,then the maximum height of the other will be:

An aircraft executes a horizontal loop with a speed of $150\, m/s$ with its wings banked at an angle of $12^\circ$. The radius of the loop is .......... $km$ $(g = 10\, m/s^2 \text{ and } \tan 12^\circ = 0.2125)$