Question

From the ground an object is vertically thrown upwards with an angle of theta.

Answer 1

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)