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JEE MAIN 2024
31.01.24 S2
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 $\frac{\sqrt{2} m u^3}{\mathrm{Xg}}$. The value of ' X ' is
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Solution
$\begin{aligned} & L=m u \cos \theta \frac{u^2 \sin ^2 \theta}{2 g} \\ & =m u^3 \frac{1}{4 \sqrt{2} g} \Rightarrow x=8\end{aligned}$
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