Rotational Motion Question 9
Question 9 - 2024 (31 Jan Shift 2)
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 $\frac{\sqrt{2} \mathrm{mu}^{3}}{\mathrm{Xg}}$. The value of ’ $\mathrm{X}^{\prime}$ is
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Answer: (8)
Solution:
$\mathrm{L}=\mathrm{mu} \cos \theta \frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$
$=m u^{3} \frac{1}{4 \sqrt{2} g} \Rightarrow x=8$
