The speed of a projectile when it is at its greatest height is √(2/5) times its speed at half the maximum height. What is its angle of projection?

Maximum height, H = (u² sin²θ) / (2g) or gH = (u² sin²θ) / 2

Velocity at highest point, v_H = u cosθ

Let v_x, v_y be the horizontal and vertical velocity of the projectile at height H/2. Then,

v_x = u cosθ and v_y² = u² sin²θ − 2g × (H/2) = u² sin²θ − gH

= u² sin²θ − (u² sin²θ)/2 = (u² sin²θ)/2

∴ Effective velocity at height H/2 = √(v_x² + v_y²)

As per question, √(2/5) × √(v_x² + v_y²) = v_H

⇒ (2/5)(v_x² + v_y²) = v_H²

⇒ (2/5)[u² cos²θ + (u²/2) sin²θ] = u² cos²θ

⇒ sin²θ = 3 cos²θ

⇒ tanθ = √3 = tan60°

∴ θ = 60°