Question 703: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

Angular momentum, $$L=p r_{\perp}=m v \cos 	heta H$$
$$$L=m v \frac{1}{\sqrt{2}} \cdot \frac{v^2 \sin ^2 45^{\circ}}{2 g} \Rightarrow L=\frac{m v^3}{4 \sqrt{2} g}$$$

Answer

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

Answer

zero

Answer

$$\frac{m v^3}{(2 g)^{1 / 2}}$$

Answer

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

Centre of Mass and Rotational Motion

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

Choose the correct answer: