Question

A body of mass $m$ is thrown with velocity $u$ from the origin of a coordinate axes at an angle with the horizon. The magnitude of the angular momentum of the particle about the origin at time $t$ when it is at the maximum height of the trajectory is proportional to

• A. μ
• B. μ2
• C. μ3
• D. independent of μ

Answer 1

The Correct Option is C

Given: A particle of mass \( m \) is projected from the origin with velocity \( u \) at an angle \( \theta \) with the horizontal.

At the maximum height, the vertical component of velocity becomes zero. The horizontal component remains:
\[ v = u \cos\theta \]
The height at the highest point is:
\[ y = \frac{u^2 \sin^2\theta}{2g} \]
At this point, the position vector \( \vec{r} \) from the origin lies along the \( x \)-\( y \) plane, and the velocity vector \( \vec{v} \) is horizontal, making a right angle with the vertical component of \( \vec{r} \).
Angular momentum about the origin is:
\[ L = |\vec{r} \times m\vec{v}| = mrv \sin\theta' \]
Here, \( \theta' = 90^\circ \), so \( \sin\theta' = 1 \).
\[ L = m \cdot y \cdot u\cos\theta = m \cdot \left( \frac{u^2 \sin^2\theta}{2g} \right) \cdot u\cos\theta = \frac{m u^3 \sin^2\theta \cos\theta}{2g} \]

Final Answer: Angular momentum is proportional to \( \mu3 = m \).