At a height of 2 meters, the kinetic potential energy is the same. With what initial speed the body was thrown?

Initial data: h (height at which the potential energy of the body is equal to the kinetic one) = 2 m.
Reference data: g (acceleration due to gravity) ≈ 10 m / s2.
1) Determine the square of the body’s velocity at a height of 2 m: m * V ^ 2/2 = m * g * h, whence V2 = 2g * h = 2 * 10 * 2 = 40 m2 / s2.
2) Calculate the speed with which the body was thrown: h = (V0 ^ 2 – V ^ 2) / 2g, whence V02 = V ^ 2 + 2g * h and V0 = √ (V ^ 2 + 2g * h) = √ ( 40 + 2 * 10 * 2) = 8.94 m / s.
Answer: The body was thrown with an initial speed of 8.94 m / s.