A projectile of mass 2 kg is launched from a point on the y -axis with an initial position vector r ₀ = 10 j m . Its initial velocity is u = (15 i + 20 j ) m s ⁻¹ . Taking the acceleration due to gravity as g = 10 m s ⁻² along the negative y -axis, the magnitude of the angular momentum of the projectile about the origin at the instant it reaches its maximum height is (in N m s )

Step-by-step solution

The projectile reaches its maximum height when its vertical velocity component becomes zero. Using the first equation of motion for the y -direction: v_y = u_y - gt 0 = 20 - 10t t = 2 s At t = 2 s , the position of the projectile is: x = u_x t = 15 2 = 30 m y = y₀ + u_y t - 1 2 gt^2 = 10 + 20(2) - 1 2 (10)(2)^2 = 10 + 40 - 20 = 30 m So, the position vector at maximum height is r = (30 i + 30 j ) m . At maximum height, the velocity vector has only the horizontal component: v = v_x i = 15 i m s ⁻¹ The angular momentu