A particle of mass 100 g is projected at time t = 0 with a speed 20 m s – 1 at an angle 45 ° to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t = 2 s is found to be K kg m 2 s - 1 . The value of K is ______. (Take g = 10 m s - 2 )

Step-by-step solution

As torque due to gravitational force will be variable, therefore we can use Δ L = ∫ 0 t τ d t . Now, torque of m g at time t = v x t m g and horizontal velocity of the projectile will remain constant v x = v cos 45 ° = 20 × 1 2 = 10 2 . L 0 = ∫ 0 2 m g v x t d t = m g v x t 2 2 = ( 0 . 1 ) ( 10 ) ( 10 2 ) 2 2 2 = 20 2 = 800   kg   m 2   s - 1 Therefore, K = 800 .