Question

A particle of mass m is projected with velocity v making an angle of ${45}^o$ with the horizontal. The magnitude of the angular momentum of the projectile about the axis of projection when the particle is at maximum height is

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

Answer 1

Answer: (B)

$\frac{m{v}^3}{4\sqrt{2}g}$

Maximum height by particle, $h=\frac{v_{vertical}^2}{2g}$

$h=\frac{v^2}{4g}$ (since , $v_{vertical}=vsin\left(45\right)=\frac{v}{\sqrt{2}}$)

Velocity at highest point $=\frac{v}{\sqrt{2}}$ {In the horizontal direction}

Magnitude of angular momentum about projectile axis $=mvh$
$=\frac{m{v}^3}{4\sqrt{2}g}$