$A$ particle moving in $x-y$ plane starts from the origin at $t=0$ with an initial velocity $(-\hat{i}+\hat{j}) \,ms^{-1}$ and undergoes an acceleration of $(6 \hat{i}+4 \hat{j}) \,ms^{-2}$. Its displacement after $2 \,s$ is (in $\,m$)

Two particles are projected from a tower in opposite directions horizontally with speed $10\,m/s$ each. At $t=1\,s$,match the following two columns.

Column $I$	Column $II$
$(A)$ Relative acceleration between two	$(p)$ $0$ $SI$ unit
$(B)$ Relative velocity between two	$(q)$ $5$ $SI$ unit
$(C)$ Horizontal distance between two	$(r)$ $10$ $SI$ unit
$(D)$ Vertical distance between two	$(s)$ $20$ $SI$ unit

The figure shows the velocity and the acceleration of a point-like body at the initial moment of its motion. The direction and the absolute value of the acceleration remain constant. Find the time when the speed becomes minimum. (Given: $a = 4\, m/s^2$,$v_0 = 40\, m/s$,$\phi = 143^o$)

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.

$Assertion$ : When a particle moves in a circle with a uniform speed,its velocity and acceleration both change.
$Reason$ : The centripetal acceleration in circular motion is dependent on the angular velocity of the body.

$A$ ball is projected at an angle of $45^{\circ}$ with the horizontal. It passes through a wall of height $h$ at a horizontal distance $d_1$ from the point of projection and strikes the ground at a distance $d_1+d_2$ from the point of projection. Then $h$ is: