A particle of mass m is projected in the x - y plane such that its trajectory is given by the equation y = x - x^2 , where and are positive constants and y is the vertical upward direction. The magnitude of the angular momentum of the particle about the origin when it is at its maximum height is (where g is the acceleration due to gravity)

Step-by-step solution

The equation of trajectory is given by y = x - x^2 . Comparing this with the standard equation of projectile motion y = x - g x^2 2 u^2 ^2 , we get: = g 2 u^2 ^2 = u = g 2 The horizontal component of velocity is constant throughout the motion and equals v_x = u = g 2 . At the maximum height, the vertical component of velocity is zero, so the velocity vector is purely horizontal: v = v_x i . The maximum height H can be found by maximizing y . For y = x - x^2 , the maximum occurs at x = 2 . Substituting this value, H