JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 22)
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
8
Explanation
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$$\begin{aligned} & \mathrm{L}=\mathrm{mu} \cos \theta \frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \\ & =\mathrm{mu}^3 \frac{1}{4 \sqrt{2} \mathrm{~g}} \Rightarrow \mathrm{x}=8 \end{aligned}$$
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