Question

Particle of mass m is projected with a velocity v₀ making an angle of 45⁰ with horizontal. The magnitude of angular momentum of the projectile about the point of projection at its maximum height is

• A. Zero
• B. $\frac{mv^{3}}{\sqrt{2}g}$
• C. $\frac{mv_{0}^{2}}{4\sqrt{2}g}$
• 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 = v₀ cos θ₀ = v₀ cos 45⁰ = $\frac{v_{0}}{\sqrt{2}}$

The angular momentum the particle about O

= L = mvr sin θ where θ = angle between $\overrightarrow{v}\&\overrightarrow{r}$

⇒ L = mv h ($\because$r sin θ = h)

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