Question

A body of mass $5kg$ is projected with a velocity $u$ at an angle ${45}^{\circ }$ to the horizontal. The magnitude of angular momentum of the body when it is at the highest position with respect to the point of projection is:

• A. $\frac{u^3}{5\sqrt{5}g}$
• B. $\frac{{5u}^3}{4\sqrt{2}g}$
• C. $\frac{u^3}{5g}$
• D. $\frac{u^3}{\sqrt{5}g}$

Answer 1

The correct option is B $\frac{{5u}^3}{4\sqrt{2}g}$

The angular momentum of body at its highest position with respect to point of projection=$m{v}_{x}H$

where $v_x$ is the horizontal component of velocity

$H$ is the maximum height the body reaches.

$v_x=ucos{45}^{\circ }=\frac{u}{\sqrt{2}}$

Also $0^2-v_y^2=-2gH$

$⟹H=\frac{v_y^2}{2g}=\frac{u^{2}si{n}^2\theta }{2g}=\frac{u^2}{4g}$

Thus Angular momentum=$m\times \frac{u}{\sqrt{2}}\times \frac{u^2}{4g}$

$=\frac{5u^3}{4\sqrt{2}g}$