Question

A particle of mass $m$ is projected from the ground with an initial speed $u$ at an angle $\alpha$, The magnitude of its angular momentum at the highest point of its trajectory about the point of projection is $\frac{m{u}^3\cos \alpha {\sin }^2\alpha }{xg}$. Find x.

Answer 1

Refer figure, vertical distance, of particle from point of origin, when at highest point of its trajectory is $H=(usin\alpha {)}^2/2g$.

The velocity at highest point is $v_h=ucos\alpha$.

So the angular momentum is $m{v}_h\times H=m{u}^{3}cos\alpha si{n}^2\alpha /2g$

so, $x=2$