A particle of mass m is projected with a velocity v making an angle of 45° with the…
Physics · JEE Advanced · NTA Exams — Rotational Motion
A particle of mass m is projected with a velocity v making an angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at maximum height h is
- zero
- \(\frac{m v^{3}}{4 \sqrt{2} g}\)
- \(\frac{m v^{3}}{\sqrt{2} g}\)
- \(\frac{m}{\sqrt{2} g v^{3}}\)
Answer
(B) m v^ 3 4 2 g
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