Question

A particle is projected with a speed v at 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

• A. zero
• B. $\frac{mvh^2}{\sqrt{2}}$
• C. $\frac{mv^2 \,h}{2}$
• D. $\frac{mvh}{\sqrt{2}}$

Answer 1

Answer: (D)

When a particle is projected with a speed v at 45$^{\circ}$ with the horizontal then velocity of the projectile at maximum height
$v'=v \,cos 45 ^{\circ}$=$\frac{v}{\sqrt{2}}$
Angular momentum of the projectile about the point of projection = mv' h
$=m\frac{m}{\sqrt{2}}h=\frac{mvh}{\sqrt{2}}$