The velocity of a particle moving in a curvilinear path in a horizontal $X-Y$ plane varies with time as $\vec{v} = (2t\hat{i} + t^2\hat{j}) \ m/s$. Here,$t$ is in seconds. At $t = 1 \ s$:

The projectile motion of a particle of mass $5\, g$ is shown in the figure. The initial velocity of the particle is $5 \sqrt{2}\, m/s$ and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points $A$ and $B$ is $x \times 10^{-2}\, kg \cdot m/s$. The value of $x$,to the nearest integer,is ...... .

$A$ ball is dropped from a height of $49\,m$. The wind is blowing horizontally. Due to wind,a constant horizontal acceleration is provided to the ball. Choose the correct statement$(s)$. (Take $g = 9.8\,m/s^2$)

$A$ particle starts from the origin at $t=0$ with a velocity $5 \hat{i} \text{ m/s}$ and moves in the $x-y$ plane under the action of a force which produces a constant acceleration of $(3 \hat{i} + 2 \hat{j}) \text{ m/s}^2$. If the $x$-coordinate of the particle at that instant is $84 \text{ m}$,then the speed of the particle at this time is $\sqrt{\alpha} \text{ m/s}$. The value of $\alpha$ is . . . . . . .

$A$ particle is projected from the ground with velocity $u$ at an angle $\theta$ with the horizontal. The horizontal range,maximum height,and time of flight are $R, H$,and $T$ respectively. They are given by $R = \frac{u^2 \sin 2\theta}{g}$,$H = \frac{u^2 \sin^2 \theta}{2g}$,and $T = \frac{2u \sin \theta}{g}$. Now,keeping $u$ fixed,$\theta$ is varied from $30^o$ to $60^o$. Then,

$A$ projectile of mass $200 \ g$ is launched in a viscous medium at an angle $60^{\circ}$ with the horizontal,with an initial velocity of $270 \ m/s$. It experiences a viscous drag force $\vec{F} = -c \vec{v}$,where the drag coefficient $c = 0.1 \ kg/s$ and $\vec{v}$ is the instantaneous velocity of the projectile. The projectile hits a vertical wall after $2 \ s$. Taking $e = 2.7$,the horizontal distance of the wall from the point of projection (in $m$) is