Question

A particle of mass m is projected with velocity v making an angle of ${45}^{\circ }$ with the horizontal. Themagnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity)

• A. $Zero$
• B. $\frac{m{v}^3}{\left(4\sqrt{2}g\right)}$
• C. $\frac{m{v}^3}{\left(\sqrt{2}g\right)}$
• D. $\frac{m{v}^2}{2g}$

Answer 1

The correct option is B

$\frac{m{v}^3}{\left(4\sqrt{2}g\right)}$

$L=\frac{m{u}^{3}cos\theta si{n}^2\theta }{2g}=\frac{m{v}^3}{\left(4\sqrt{2}g\right)}[As\theta ={45}^{\circ }]$