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 its apex, the object's vertical velocity is zero; only its horizontal component, \( u_x = u \cos 45^\circ = \frac{u}{\sqrt{2}} \), persists.

The maximum altitude \( h \) achieved is calculated as:

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

The angular momentum \( L \) relative to the launch point at the apex is:

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

Consequently, the value of \( X \) is determined to be:

\[X = 8. \]