Question

A particle of mass m is projected with velocity V making an angle of \(45^{\circ }\) with the horizontal. The magnitude of the angular momentum of the projectile about an axis of projection when the particle is at maximum height h is (\(g\) = acceleration due to gravity) \(\left(\sin 45^{\circ }=\frac{1}{\sqrt{2}}\right)\) \(\left(\cos 45^{\circ }=\frac{1}{\sqrt{2}}\right)\)

• A. zero
• B. \(\frac{mV^{3}}{4\sqrt{2}g}\)
• C. \(\frac{mV^{3}}{\sqrt{2}g}\)
• D. \(m\sqrt{2gh^{2}}\)

Hint:

At the top only the horizontal velocity remains, and the angular momentum about the point of projection equals m times that velocity times the height.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
At maximum height the velocity is purely horizontal: \(v_{x}=V\cos45^{\circ}=\dfrac{V}{\sqrt{2}}\).
Angular momentum about the point of projection is \(L=m\,v_{x}\,h\) because the perpendicular distance of the line of motion is \(h\).

Step 2: Find the Maximum Height
\(h=\dfrac{V^{2}\sin^{2}45^{\circ}}{2g}=\dfrac{V^{2}}{4g}\).

Step 3: Calculation
\(L=m\cdot\dfrac{V}{\sqrt{2}}\cdot\dfrac{V^{2}}{4g}=\dfrac{mV^{3}}{4\sqrt{2}\,g}\).

Final Answer:
The angular momentum is mV^3 divided by 4sqrt2 g. \[ \boxed{\text{(B) }\dfrac{mV^{3}}{4\sqrt{2}g}} \]