$A$ bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 \; km$ away. By adjusting its angle of projection,can one hope to hit a target $5.0 \; km$ away? Assume the muzzle speed to be fixed,and neglect air resistance.

Fill in the blanks:
$(a)$ For vectors $\vec{A} = 3\hat{i} + 2\hat{j}$ and $\vec{B} = \hat{i} + \hat{j} - 2\hat{k}$,the $y$-component of the vector $(\vec{A} - \vec{B})$ is .....
$(b)$ The formula for the total time of flight of a projectile projected vertically upwards with an initial velocity $u$ is ..........
$(c)$ The formula for the centripetal acceleration of a particle moving in a circular path of radius $R$ with a constant angular velocity $\omega$ is ......
$(d)$ The component of any vector is always .........

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 trajectory of a particle in projectile motion is given by $y = x - \frac{x^2}{80}$. Here, $x$ and $y$ are in meters. For this projectile motion, match the following with $g = 10 \, m/s^2$.

$Column-I$	$Column-II$
$(A)$ Angle of projection	$(p)$ $20 \, m$
$(B)$ Angle of velocity with horizontal after $4 \, s$	$(q)$ $80 \, m$
$(C)$ Maximum height	$(r)$ $45^{\circ}$
$(D)$ Horizontal range	$(s)$ $\tan^{-1}(1/2)$

The initial velocity of a projectile is $\vec{u} = (4\hat{i} + 3\hat{j})\,m/s$. It is moving with uniform acceleration $\vec{a} = (0.4\hat{i} + 0.3\hat{j})\,m/s^2$. The magnitude of its velocity after $10\,s$ is.........$m/s$.

$A$ body of mass $0.5 \,kg$ is projected under gravity with a speed of $98 \,m/s$ at an angle of $30^\circ$ with the horizontal. The change in momentum (in magnitude) of the body is ......... $N-s$.