A particle of mass $$m$$ is projected with a velocity $$\upsilon $$ making an angle of $${30^ \circ }$$ with the horizontal. The magnitude of $$\left( {{V_h} \times h} \right)$$  of the projectile when the particle is at its maximum height $$h$$

Solution :
$${V_h} = V\cos \theta $$
where $$h$$ is the maximum height
$$\eqalign{ & {V_h} \times h = \left( {\upsilon \cos \theta } \right)\left( {\frac{{{\upsilon ^2}{{\sin }^2}\theta }}{{2g}}} \right) \cr & = \frac{{{\upsilon ^3}{{\sin }^2}\theta \cos \theta }}{{2g}} \cr & = \frac{{\sqrt 3 {\upsilon ^3}}}{{16\,g}} \cr} $$