8) A soccer goal is 2.44 m high. A player kicks the ball at a distance 10.0 m from the goal at an angle of 25.0°. The ball hits the crossbar at the top of the goal. What is the initial speed of the soccer ball?

The initial speed of the soccer ball is 16.38 m/s

Explanation:

given;

vertical distance y = 2.44 m

horizontal distance x = 10.0 m

angle of projection θ = 25.0°

Initial velocity has two components, Vₓ and V

Vₓ  = V cosθ

V = V sinθ

The horizontal distance = x = Vₓt + ¹/₂ ˣ g ˣ t², but g =0

x = Vₓt =  V cosθ *t

10 = V cos25 *t

10 = 0.906V*t

V*t = 10/0.906 = 11.038 m

The vertical distance (g = - g, because it upward motion against gravity)

y = V*t -¹/₂ ˣ g ˣ t²

2.44 = (V sinθ)t - ¹/₂ ˣ 9.8 ˣ t²

2.44 = (V*t)sinθ - ¹/₂ ˣ 9.8 ˣ t²

2.44 = (11.038)sin25° - 4.9t²

2.44 = (11.038)*0.4226 - 4.9t²

2.44 = 4.6647 - 4.9t²

4.9t² = 4.6647 - 2.44 = 2.2247

t² = 2.2247/4.9

t² = 0.454

t = √0.454

t = 0.674 s

Recall that V*t = 11.038 m

V*0.674 = 11.038 m, solve for V

V = 11.038/0.674

V = 16.38 m/s

Therefore, the initial speed of the soccer ball is 16.38 m/s