Question

A body of mass 'm' is projected with a speed 'u' making an angle of $45^\circ$ with the ground. The angular momentum of the body about the point of projection, at the highest point, is expressed as $\frac{\sqrt{2} \, m u^3}{X g}$. The value of 'X' is ______.

Answer 1

Correct Answer: 8

At the highest point, the vertical component of the velocity becomes zero, and only the horizontal component $u_x = u \cos 45^\circ = \frac{u}{\sqrt{2}}$ remains.

The maximum height $h$ reached by the body is given by:

$$h = \frac{(u \sin 45^\circ)^2}{2g} = \frac{\left(\frac{u}{\sqrt{2}}\right)^2}{2g} = \frac{u^2}{4g}.$$

The angular momentum $L$ about the point of projection at the highest point is:

$$L = m \cdot u_x \cdot h = m \cdot \frac{u}{\sqrt{2}} \cdot \frac{u^2}{4g} = \frac{\sqrt{2}mu^3}{8g}.$$

Thus, the value of $X$ is:

$$X = 8.$$