The boy hit a soccer ball, giving him a speed of 20 m / s. The ball hit the wall of the house at a height of 7.2 m

The boy hit a soccer ball, giving him a speed of 20 m / s. The ball hit the wall of the house at a height of 7.2 m, and at the moment of impact, its speed was directed horizontally. How far from the wall of the house was the ball originally located? At what angle to the horizon was the initial velocity of the ball directed? What is the modulus of its movement?

V0 = 20 m / s.

h = 7.2 m.

g = 10 m / s2.

L -?

∠α -?

S -?

Horizontally, the ball flies uniformly at a speed V0 * cosα, vertically with an acceleration of gravity g and an initial speed V0 * sinα until it comes to a complete stop.

L = V0 * cosα * t.

h = (V0 * sinα) ^ 2/2 * g.

sin2α = 2 * g * h / V0 ^ 2.

∠α = arcsin (√ (2 * g * h) / V0.

∠α = arcsin (√ (2 * 10 m / s2 * 7.2 m) / 20 m / s = arcsin0.6 = 37 °.

t = V0 * sinα / g.

t = 20 m / s * sin37 ° / 10 m / s2 = 1.2 s.

L = 20 m / s * cos37 ° * 1.2 s = 18.96 m.

S = √ (L2 + h2).

S = √ ((18.96 m) 2 + (7.2 m) 2) = 20 m.

Answer: the impact speed directed at an angle ∠α = 37 °, L = 18.96 m, the movement of the ball was S = 20 m.