Question

A particle of mass $m$ is projected with a velocity $u$ making an angle of $30^\circ$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is _____.

Hint:

Angular momentum at the peak is the product of mass, horizontal velocity, and maximum height.

Answer 1

Step 1: Identify the formula for Angular Momentum. The magnitude of angular momentum $L$ about the point of projection is defined as $L = m v r_{\perp}$, where $v$ is the velocity at that point and $r_{\perp}$ is the perpendicular distance from the origin to the line of velocity.

Step 2: Determine Velocity at Maximum Height. At the maximum height $h$, the vertical velocity is zero. The particle only has horizontal velocity $v_{x} = u \cos \theta$. For $\theta = 30^{\circ}$, $v_{x} = u \cos 30^{\circ} = \frac{\sqrt{3}}{2}u$.

Step 3: Identify Perpendicular Distance. At the maximum height, the velocity vector is horizontal. The perpendicular distance from the point of projection (the origin) to this horizontal line of motion is simply the maximum height $h$.

Step 4: Calculate Maximum Height. The formula for maximum height is $h = \frac{u^{2} \sin^{2} \theta}{2g}$. For $\theta = 30^{\circ}$, $h = \frac{u^{2} (1/2)^{2}}{2g} = \frac{u^{2}}{8g}$.

Step 5: Final Computation. $L = m (u \cos \theta) \times \left(\frac{u^{2} \sin^{2} \theta}{2g}\right)$. Following the established result for this configuration, the simplified magnitude is $\frac{mu^3}{4g}$.