At a height of 4.4 m from the surface of the Earth, the ball had a speed of 10 m / s.

At a height of 4.4 m from the surface of the Earth, the ball had a speed of 10 m / s. How fast will the ball move at the surface of the Earth?

h0 = 4.4 m.

V0 = 10 m / s.

g = 10 m / s2.

V -?

To solve the problem, we will use the law of conservation of total mechanical energy. The total mechanical energy of the ball remains unchanged Ep = const.

En0 + Ek0 = Ek.

En0 = m * g * h0.

Ek0 = m * V0 ^ 2/2.

Ek = m * V ^ 2/2.

m * g * h0 + m * V0 ^ 2/2 = m * V2 / 2.

g * h0 + V0 ^ 2/2 = V2 / 2.

V2 = 2 * g * h0 + V0 ^ 2.

V = √ (2 * g * h0 + V0 ^ 2).

V = √ (2 * 10 m / s2 * 4.4 m + (10 m / s) ^ 2) = 13.7 m / s.

Answer: at the moment the ball touches the ground, it will have a speed of V = 13.7 m / s.