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 particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity)

• A. Zero
• B. mv³/$(4\sqrt{2}g)$
• C. mv³/$(\sqrt{2}g)$
• D. mv²/2g

Answer 1

Answer: (B)

mv³/$(4\sqrt{2}g)$

Answer 2

$L = \frac{mu^{3}\cos\theta\sin^{2}\theta}{2g}$= $\frac{mv^{3}}{(4\sqrt{2}g)}$ [As θ = 45^o]