The body was thrown vertically upward at a speed of 20 m / s.

The body was thrown vertically upward at a speed of 20 m / s. At what height from the initial level will its kinetic energy be two times less than its potential energy?

Task data: V0 (body lifting speed) = 20 m / s; En1 (potential energy at the desired height) = 2Ek1 (kinetic energy at the desired height).

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

To find the height that the body must reach under the given conditions, we apply the law of conservation of energy: Ek0 = Ek1 + En1 = 1.5En1; m * V0 ^ 2/2 = 1.5 * m * g * hx, whence we express: hx = V0 ^ 2 / (2 * 1.5 * g) = V0 ^ 2 / (3 * g).

Calculation: hx = 20 ^ 2 / (3 * 10) = 13.33 m.

Answer: The body should be at a height of 13.33 m.