$A$ 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

$A$ projectile is thrown into air with velocity $15 \ m/s$ at an angle $30^{\circ}$ with the horizontal. After what time is its direction of motion perpendicular to its initial direction (in $s$)? (Assume $g = 10 \ m/s^2$)

$A$ point traversed $3/4$th of the circle of radius $R$ in time $t$. The magnitude of the average velocity of the particle in this time interval is

$A$ projectile is launched with an initial speed of $40 \ m \ s^{-1}$ at an angle of $30^{\circ}$ above the ground. The projectile lands on a hillside $2.0 \ s$ later. The net displacement from where the projectile was launched to where it hits the target is (Take $g = 10 \ m \ s^{-2}$)

If $\alpha$ is angular acceleration,$\omega$ is angular velocity,and $a$ is the centripetal acceleration,then which of the following is true?

$A$ particle performs uniform circular motion with an angular momentum $L$. If the frequency of the particle's motion is doubled and its kinetic energy is halved, then its angular momentum becomes