A body weighing 1 kg is thrown vertically upward at a speed of 30 m / s

A body weighing 1 kg is thrown vertically upward at a speed of 30 m / s Determine the kinetic and potential energy of the body 2 seconds after the throw

Data: m (thrown body weight) = 1 kg; Vн (throwing speed) = 30 m / s; t (considered travel time) = 2 s.

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

1) Kinetic energy: Ek = m * Vk ^ 2/2 = m * (Vn – g * t) ^ 2/2 = 1 * (30 – 10 * 2) ^ 2/2 = 50 J.

2) Potential energy: Ep = m * g * h = m * g * (Vn * t – g * t ^ 2/2) = 1 * 10 * (30 * 2 – 10 * 2 ^ 2/2) = 400 J.

Check: Ek0 = m * Vn ^ 2/2 = Ek + Ep.

1 * 30 ^ 2/2 = 50 + 400.

450 = 450 (correct).

Answer: After 2 seconds, the body will have a kinetic energy of 50 J and a potential of 400 J.