Question

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

Find the mean velocity between the start point and the maximum height.. choose the correct answer and show all the steps. plzz help...​

Answer 1

u/2 √(1 + 3 cos² θ)

Step-by-step 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)