$A$ body of mass '$m$' is projected with a speed '$u$' making an angle of $45^{\circ}$ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $\frac{\sqrt{2} mu^3}{Xg}$. The value of '$X$' is

The kinetic energy $K$ of a particle moving along a circle of radius $R$ depends on the distance covered $s$ as $K = as^2$,where $a$ is a constant. The force acting on the particle is:

$Assertion$ : 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.

$A$ 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.

Consider the two statements related to circular motion in usual notations:
$A$. In uniform circular motion,$\vec{\omega}$,$\vec{v}$,and $\vec{a}$ are always mutually perpendicular.
$B$. In non-uniform circular motion,$\vec{\omega}$,$\vec{v}$,and $\vec{a}$ are always mutually perpendicular.

$A$ 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 $\,m$)