A particle of mass m projected with a velocity ‘u’ making an angle of 30º with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is :
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Question

A particle of mass m projected with a velocity ‘u’ making an angle of 30º with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is :

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G1 Physics · Accepted Answer

At maximum height, the vertical component of velocity is zero. Horizontal velocity: vₓ = u cosθ Maximum height: H = u² sin²θ /(2g) Angular momentum about the point of projection: L = m vₓ H = μ \mu μ cosθ × [u² sin²θ /(2g)] For θ = 30º: cos30º = √3/2 sin30º = 1/2 Therefore, L = ( μ \mu μ ³/2g) × (√3/2) × (1/2)² = √3mu³/(16g)