The body is thrown vertically downward at a speed of 10 m / s from a height of 30 m.

The body is thrown vertically downward at a speed of 10 m / s from a height of 30 m. At what height from the earth’s surface the kinetic energy will double in comparison with the initial one?

Data: Vн (throwing speed, initial speed) = 10 m / s; hн (initial height) = 30 m; g (acceleration due to gravity) ≈ 10 m / s2.

To determine the desired height of the specified body, we use the law of conservation of energy: Ek0 + Ep0 = 2Ek0 + Ep; Ep0 = Ek0 + Ep; m * g * hн = m * Vn ^ 2/2 + m * g * h, whence we express: h = (g * hn – Vn ^ 2/2) / g = (10 * 30 – 10 ^ 2/2) / 10 = 25 m.

Answer: The kinetic energy of the specified body will double at a height of 25 m.