A ball of mass 160 g is thrown up at an angle of 60 ° to the horizontal at a speed of 10 m s - 1 . The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly g = 10 m s - 2

Step-by-step solution

The angular momentum L → = r → × p → At maximum height angular momentum about O , L = m   v cosθ   H = m v cosθ × u 2 sin 2 θ 2 g                 f o r   h o r i z o n t a l   v = u = m u 3 sin 2 60 ° cos 60 ° 2 g = 3   kg   m 2   s - 1