Body mass projected with speed making

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

$\begin{aligned} & L=m u \cos \theta \frac{u^2 \sin ^2 \theta}{2 g} \\ & =m u^3 \frac{1}{4 \sqrt{2} g} \Rightarrow x=8\end{aligned}$

Question Tags

JEE Main

Physics

Easy

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