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 its maximum height h, are:
- zero
- \(\frac{M v^{3}}{4 \sqrt{2} g}\)
- \(M v^{3} / \sqrt{2} g\)
- \(m \sqrt{\left(2 g h^{3}\right)}\)
Answer
(B) M v^ 3 4 2 g, (D) m (2 g h^ 3 )
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