A projectile is projected at 30^ from horizontal with initial velocity 40 ~ms ⁻¹ . The velocity of the projectile at t =2 ~s from the start will be: (Given g =10 ~m / s ^2 )

Step-by-step solution

Given, Initial velocity of projectile, u=40 ~m / s Angle, =30^ Time of flight T = 2 u g = 2 40 1 10 2 =45 ( g =10 ~m / s ^2 ) It means projectile is at maximum height at t=2 ~s . At maximum height vertical component of velocity is zero. aligned & Velocity at t =2 ~s = V _ x = u =40 30^ & =20 3 ~ms ⁻¹ . & aligned