A ball is thrown at an angle of 60^ above the horizontal. If its speed at the maximum height is 15 m s ⁻¹ , what is its initial speed of projection?

Step-by-step solution

At the highest point of a projectile's trajectory, the vertical component of velocity is zero. The speed at this point is entirely due to the horizontal component of the initial velocity, which remains constant throughout the motion. Let the initial speed be u . The speed at the maximum height is given by: v = u Given that v = 15 m s ⁻¹ and = 60^ : 15 = u 60^ 15 = u 1 2 u = 30 m s ⁻¹ Answer: 30 m s ⁻¹