A body of mass 'm' is projected with a speed 'u' making an angle of 45° with the ground. The angular momentum of the body about the point of projection, at the highest point, is expressed as . The value of ‘X’ is

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Solution

Q.
A body of mass 'm' is projected with a speed 'u' making an angle of 45° 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 _____. [2024]

Ans.
(8)

$L = m u \cos θ \frac{u^{2} \sin^{2} θ}{2 g} = m u^{3} \frac{1}{4 \sqrt{2} g} \Rightarrow x = 8$

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