A particle is projected from the ground with a speed of 40 m s ⁻¹ at an angle of 30^ with the horizontal. What is the magnitude of the change in its velocity between the point of projection and the highest point of its trajectory?

Step-by-step solution

Let the point of projection be the origin. The initial velocity vector u can be resolved into horizontal and vertical components: u = u i + u j u = 40 30^ i + 40 30^ j = 20 3 i + 20 j m s ⁻¹ At the highest point of the trajectory, the vertical component of velocity becomes zero, while the horizontal component remains constant. Velocity at the highest point, v = 20 3 i m s ⁻¹ The change in velocity vector is v = v - u v = (20 3 i ) - (20 3 i + 20 j ) = -20 j m s ⁻¹ The magnitude of the change in velocity is | v | =