Question

A particle is projected with a speed $v$ at $45 ^\circ $ with the horizontal. The magnitude of angular momentum of the projectille about the point of projection when the particle is at its maximum height $h$ is

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

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.
$v^{′}=vcos 45 ^\circ = \frac{v}{\sqrt{2}}$
Angular momentum of the projectile about the point of projection
$=mv′h$
$=m\frac{v}{\sqrt{2}}h=\frac{m v h}{\sqrt{2}}$