The body was thrown vertically upward with an initial speed of 40 m / s. At what height is the kinetic

The body was thrown vertically upward with an initial speed of 40 m / s. At what height is the kinetic energy twice the potential

To calculate the height at which a given body is required to be, we apply the law of conservation of energy: Ek0 = Ek1 + Ep1 = 2En1 + En1 = 3En1 and m * V02 / 2 = 3m * g * hx, whence we express: hx = V02 / 6g.

Constants and variables: V0 – throw speed (V0 = 40 m / s); g – acceleration due to gravity (g ≈ 10 m / s2).

Calculation: hx = V02 / 6g = 402 / (6 * 10) = 26.67 m.

Answer: The kinetic energy of the body will be 2 times higher than the potential at an altitude of 26.67 m.