The velocity of a projectile at the initial point $A$ is $(2\hat i + 3\hat j) \text{ m/s}$. Its velocity (in $\text{m/s}$) at point $B$ is

$A$ particle is projected horizontally from a tower with velocity $10\,m/s$. Taking $g=10\,m/s^2$,match the following two columns at time $t=1\,s$.

Column $I$	Column $II$
$(A)$ Horizontal component of velocity	$(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity	$(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement	$(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement	$(s)$ $20$ $SI$ unit

$A$ particle is projected from the ground with a kinetic energy $E$ at an angle of $60^{\circ}$ with the horizontal. Its kinetic energy at the highest point of its motion will be

$A$ small object is thrown at an angle $45^{\circ}$ to the horizontal with an initial velocity $v_0$. The velocity is averaged for the first $\sqrt{2} \,s$ and the magnitude of the average velocity comes out to be the same as that of the initial velocity,i.e.,$|v_0|$. The magnitude $|v_0|$ will be (take $g=10 \,m/s^2$):

If bullets are fired in all possible directions from the same point with an equal velocity of $10 \,m \,s^{-1}$ and an angle of projection $45^{\circ}$,then the area covered by the bullets on the ground is nearly (Acceleration due to gravity $g = 10 \,m \,s^{-2}$) (in $\,m^2$)

Match Column-$I$ with Column-$II$.

Column-$I$	Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed	$(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed	$(b)$ $0^o$