$A$ 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

$A$ particle is moving with constant speed $v$ in the $xy$ plane as shown in the figure. The magnitude of its angular velocity about point $O$ is .........

$A$ car is moving with a velocity of $4 \,m \,s^{-1}$ towards east. After a time of $4 \,s$, if it is heading north-east with a velocity of $4 \sqrt{2} \,m \,s^{-1}$, then the average velocity of the car is

$A$ particle is projected from the ground with some initial velocity making an angle of $45^{\circ}$ with the horizontal. If it reaches a height of $7.5 \ m$ above the ground,while it travels a horizontal distance of $10 \ m$ from the point of projection,then the initial speed of the particle is (assume,$g=10 \ m/s^2$): (in $m/s$)

$A$ merry-go-round starts from rest and accelerates at $0.4 \ rad \ s^{-2}$ for the first $5 \ s$. It then rotates at this constant angular velocity for $30 \ s$ and finally comes to rest with the same magnitude of angular deceleration. What is the total linear distance traveled by a child sitting $3 \ m$ from the center of the merry-go-round?

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	$(s)$ $\tan^{-1}(1/2)$