A projectile is launched from the ground with an initial speed u . At the highest point of its trajectory, its speed is observed to be u 2 . What is the angle of projection of the projectile with the vertical direction?

Step-by-step solution

Let the angle of projection with the horizontal be . The horizontal component of velocity remains constant throughout the projectile motion. At the highest point, the vertical component of velocity is zero, so the total speed is just the horizontal component of the velocity. Speed at highest point, v = u Given that v = u 2 , we have: u = u 2 = 1 2 = 60^ The angle of projection with the vertical is 90^ - = 90^ - 60^ = 30^ . Answer: 30^