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Asked in JEE Main 24th Jan 1st Shift 2025 · Angular momentum of a particle
At the top, speed is horizontal vₓ=v₀ cos 45°=(v₀)/(√2) at height H=(v₀² sin² 45°)/(2g)=(v₀²)/(4g).
L=m vₓ H=m(v₀)/(√2)(v₀²)/(4g)=(mv₀³)/(4√2 g), directed into the page (-z) for motion in the +x, +y plane.
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