A small object is thrown at an angle 45^ to the horizontal with an initial velocity v ₀ . The velocity is averaged for first 2 ~s and the magnitude of average velocity comes out to be same as that of initial velocity, i.e. | v _ o | . The magnitude | v ₀ | will be (take, g=10 ~m / s ^2 )

Step-by-step solution

Let object is at B(x, y) after t= 2 ~s Then, x=u_x t=v₀ 45^ 2 =v₀ and y=u_y t- 1 2 a_y t^2=v₀ ( 45^ ) 2 - 1 2 (10) ( 2 )^2 = (v₀-10 ) m Displacement O B of particle is O B= O A^2+A B^2 = v₀^2+ (v₀-10 )^2 So, v_ avg = O B t =v₀ or O B=v₀ t array lrl & v₀^2+ (v₀-10 )^2 =v₀ 2 & v₀^2+ (v₀-10 )^2=2 v₀^2 & v₀-10= v₀ & 2 v₀=10 v₀=5 ~ms ⁻¹ array