The body was thrown vertically downward with an initial speed of 10 m / s from a height of 100 m.

The body was thrown vertically downward with an initial speed of 10 m / s from a height of 100 m. At what height will the body’s kinetic energy be equal to its potential energy?

Given:

v0 = 10 meters per second – the initial speed with which the body was thrown down;

h0 = 100 meters – the height from which the body was thrown;

g = 10 Newton / kilogram – acceleration of gravity;

Ekin = Epot – at a certain height the kinetic and potential energies of the body are equal.

It is required to determine h (meter) – the height of the body.

By the condition of the problem, we will not take into account air resistance when solving. Then, according to the law of conservation of energy:

Ekin (at height h0) + Epot (at height h0) = Ekin (at height h) + Epot (at height h);

Ekin (at the height h0) + Epot (at the height h0) = 2 * Epot (at the height h);

m * v0 ^ 2/2 + m * g * h0 = 2 * m * g * h, where m is body weight;

v0 ^ 2/2 + g * h0 = 2 * g * h;

v0 ^ 2 + 2 * g * h0 = 4 * g * h;

h = (v0 ^ 2 + 2 * g * h0) / (4 * g) = (10 ^ 2 + 2 * 10 * 100) / (4 * 10) = (100 + 2000) / 40 = 2100/40 = 210/4 = 52.5 meters.

Answer: potential and kinetic energies will be equal at a height of 52.5 meters.