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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA particle of mass \( m = 2\,\text{kg} \) is projected with an initial velocity \( u = 20\sqrt{2}\,\text{m/s} \) at an angle of \( 45^\circ \) with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when it reaches its maximum height is (take \( g = 10\,\text{m/s}^2 \)):
Correct answer · Option 2
800 J·s
Explanation
At maximum height, \( v_x = u\cos 45^\circ = 20\,\text{m/s} \), and \( H = \frac{u^2 \sin^2 45^\circ}{2g} = \frac{(800)(1/2)}{20} = 20\,\text{m} \). The perpendicular distance from the line of velocity to the point of projection is \( H \). Thus, \( l = m v_x H = 2 \times 20 \times 20 = 800\,\text{J}\cdot\text{s} \).
