Physics - Rotation Question with Solution | TestHub

At the maximum height $h$, the velocity of the particle is purely horizontal.

The horizontal component of the velocity remains constant throughout the projectile motion and is given by:

$v_x=v\cos 45^{\circ }=\frac{v}{\sqrt{2}}$

The angular momentum $L$ about the point of projection is the product of the linear momentum and the perpendicular distance from the point of projection to the line of motion.

At the maximum height, the line of motion is horizontal and its perpendicular distance from the point of projection is $h$.

Therefore, the magnitude of angular momentum is:

$L=(m{v}_x)\times h$

$L=m\left(\frac{v}{\sqrt{2}}\right)h$

$L=\frac{mvh}{\sqrt{2}}$