Question

Particle of mass m is projected with a velocity $v_{0}$ making an angle of $45^{0}$ with horizontal. The magnitude of angular momentum of the projectile about the point of projection at its maximum height is

• A. Zero
• B. $\frac{mv^{3}}{\sqrt{2g}}$
• C. $\frac{mv_{0}^{2}}{4\sqrt{2g}}$
• D. $m\sqrt{2gh^{3}}$

Answer 1

Answer: (D)

$m\sqrt{2gh^{3}}$

Answer 2

Speed of the particle at the top = horizontal component of the speed of projection

⇒$v = v_{0}\cos\theta_{0} = v_{0}\cos 45^{0} = \frac{v_{0}}{\sqrt{2}}$

The angular momentum the particle about O.

= L = mvr sinθ where

θ = angle between $\overset{\rightarrow}{v}\&\overset{\rightarrow}{r}$ ⇒ L = mvh ($\left( \because r\sin\theta = h \right)$

Putting h = $\frac{v_{0}^{2}\sin^{2}\theta}{2g} = \frac{v_{0}^{2}\sin^{2}45^{0}}{2g} = \frac{v_{0}^{2}}{4g}$we obtain

L = $\frac{mv_{0}^{3}}{4\sqrt{2}}$putting $v = \sqrt{(4gh)}$

We obtain L = $m\sqrt{2gh^{3}}$