Answer:
u/2 √(1 + 3 cos² θ)
Explanation:
The object is thrown at an angle θ, so the velocity has two components, vertical and horizontal.
Initially, the vertical component is u sin θ and the horizontal component is u cos θ.
At the maximum height, the vertical component is 0 and the horizontal component is u cos θ.
The mean vertical velocity is:
(u sin θ + 0) / 2 = u/2 sin θ
The mean horizontal velocity is:
(u cos θ + u cos θ) / 2 = u cos θ
The net mean velocity can be found with Pythagorean theorem:
v² = (u/2 sin θ)² + (u cos θ)²
v² = u²/4 sin² θ + u² cos² θ
v² = u²/4 (1 − cos² θ) + u² cos² θ
v² = u²/4 (1 − cos² θ) + u²/4 (4 cos² θ)
v² = u²/4 (1 − cos² θ + 4 cos² θ)
v² = u²/4 (1 + 3 cos² θ)
v = u/2 √(1 + 3 cos² θ)
