# A body weighing 1 kg, thrown from a tower 7 m high at a speed of 8 m / s, fell on the Earth’s surface at a speed of 14 m / s

**A body weighing 1 kg, thrown from a tower 7 m high at a speed of 8 m / s, fell on the Earth’s surface at a speed of 14 m / s, calculate the change in the total mechanical energy of the body at the moment of its fall on the Earth’s surface.**

To find the losses of the total mechanical energy of a thrown body, we use the equality: Ep0 + Ek0 = Ek1 + A, whence we express: A = Ep0 + Ek0 – Ek1 = m * g * hb + m * Vn ^ 2/2 – m * Vk ^ 2 / 2 = m * (g * hb + Vn ^ 2/2 – Vk ^ 2/2).

Variables and constants: m – body weight (m = 1 kg); g – acceleration due to gravity (g ≈ 10 m / s2); hb – tower height (hb = 7 m); Vн – initial speed (Vн = 8 m / s); Vк – falling speed (Vк = 14 m / s).

Calculation: A = m * (g * hb + Vn ^ 2/2 – Vk ^ 2/2) = 1 * (10 * 7 + 8 ^ 2/2 – 14 ^ 2/2) = 4 J.

Answer: At the moment of the fall, the total mechanical energy decreased by 4 J.