A projectile is projected at 30 ° from horizontal with initial velocity 40 m s - 1 . The velocity of the projectile at t = 2 s from the start will be:

Step-by-step solution

It is given that v = 40   m   s - 1 . The components of the velocity will be v y = v sin 30 ° = 40   m   s - 1 2 = 20   m   s - 1 v x = v cos 30 ° = 40 3 2   m   s - 1 = 20 3     m   s - 1 The time taken to reach maximum height is T = u sin 30 ° g = 20   m   s - 1 10   m   s - 2 = 2   s Hence, at T = 2   s ,   as   v y = 0   m   s - 1   therefore   v n e t = v x = 20 3   m   s