Question

A particle of mass m is projected from ground with speed u at an angle of 30° with the horizontal. Find its angular momentum about the point of projection when it reaches its maximum height.

• A. $\frac{mu^3}{16g}$
• B. ${\frac{\sqrt3mu^3}{16g}}$
• C. $\frac{mu^3}{3g}$
• D. Zero

Answer 1

Answer: (B)

${\frac{\sqrt3mu^3}{16g}}$

Answer 2

Step 1. Determine the Horizontal Component of Velocity: The horizontal component of the initial velocity $u_x = u \cos \theta$. Given $\theta = 30^\circ$:

$u_x = u \cos 30^\circ = u \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3} \, u}{2}$

Step 2. Calculate the Vertical Component of Initial Velocity: The vertical component of the initial velocity $u_y = u \sin \theta$. With $\theta = 30^\circ$:

$u_y = u \sin 30^\circ = u \times \frac{1}{2} = \frac{u}{2}$

Step 3. Find Time to Reach Maximum Height: At maximum height, the vertical velocity becomes zero. Using $v_y = u_y - gt$:

$0 = \frac{u}{2} - gt \Rightarrow t = \frac{u}{2g}$

Step 4. Calculate Maximum Height $H$: Use the equation $H = u_y t - \frac{1}{2}gt^2$:

$H = \frac{u}{2} \times \frac{u}{2g} - \frac{1}{2} g \left( \frac{u}{2g} \right)^2 = \frac{u^2}{4g} - \frac{u^2}{8g} = \frac{u^2}{8g}$

Step 5. Calculate Angular Momentum about the Point of Projection: Angular momentum $L$ about the point of projection is given by $L = mu_x H$. - Substituting $u_x = \frac{\sqrt{3} \, u}{2}$ and $H = \frac{u^2}{8g}$:

$L = m \times \frac{\sqrt{3} \, u}{2} \times \frac{u^2}{8g} = \frac{\sqrt{3} \, mu^3}{16g}$