A particle is projected from the ground at an angle of 60^ with the horizontal. It strikes a vertical wall situated at a horizontal distance of 20 3 m from the point of projection after 2 s. The height from the ground at which the particle strikes the wall is (Take g = 10 m/s ^2 )

Step-by-step solution

The horizontal distance covered in time t = 2 s is x = 20 3 m. The horizontal component of initial velocity is: u_x = x t = 20 3 2 = 10 3 m/s The vertical component of initial velocity is related to the horizontal component by the angle of projection = 60^ : u_y = u_x 60^ = 10 3 3 = 30 m/s The height H at which the particle strikes the wall is the vertical displacement at t = 2 s: H = u_y t - 1 2 gt^2 H = 30(2) - 1 2 (10)(2)^2 H = 60 - 20 = 40 m Answer: 40 m