Question 202 of 280

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

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Answer: 8

Consider projection as shown in above diagram.
At the topmost point, velocity of the body will be $u\cos 45^{\circ}$.
Therefore, we can write the required angular momentum as
$L=\left(m{u}_{H}\right)\times H =\left(mu\cos \theta \right)\left(\dfrac{{u}^{2}{\sin }^{2}\theta }{2g}\right)$
$=m{u}^{3}\times \dfrac{1}{4\sqrt{2}g} \Rightarrow X=8$