# A stone weighing 400 g was thrown at a speed of 20 m / s in a horizontal direction from a tower whose height is 50 m.

**A stone weighing 400 g was thrown at a speed of 20 m / s in a horizontal direction from a tower whose height is 50 m. Find the potential and kinetic energy of the stone 2 s after the start of its movement.**

Let’s find the potential energy in 2 seconds. According to the formula known in physics, for a freely falling body.

H = (G * t ^ 2): 2 = (9.8 * 2 ^ 2): 2 = (9.8 * 4): 2 = 19.6. In 2 seconds the body flew 19.6 m. This means that its height above the ground is 50 m – 19.6 m – 30.4 m. The potential energy of the body is equal.

E = M * G * H = 0.4 kg * 9.8 * 30.6 = 119.952 J.

Now let’s find the kinetic energy of the body. The horizontal component is 20 m / s, the vertical component is V = G * t = 9.8 * 2 = 19.6 m / s. The sum of the speeds is equal. V ^ 2 = 20 ^ 2 + 19.6 ^ 2 = 384.16 + 400 = 784.16. Hence V = 19.6 m / s.

The kinetic energy of the body is equal. E = 0.4 * 784.16: 2 = 156.832 J.