# A ball with a mass of 100 g falls from a certain height with zero initial velocity.

**A ball with a mass of 100 g falls from a certain height with zero initial velocity. Its kinetic energy when falling to the Earth is 6 J, and the energy loss due to air resistance was 20: the total mechanical energy at the top point of the trajectory. From what height did the ball fall?**

Given:

m = 100 grams = 0.1 kilograms – the mass of the ball;

Eк = 6 J is the kinetic energy of the ball at the surface of the earth;

g = 10 m / s ^ 2 – acceleration of gravity;

n = 20% = 0.2 is the coefficient of energy loss from the total mechanical energy at the upper point of the trajectory.

It is required to determine H (meters) – the height from which the ball fell.

For this case, the law of conservation of energy will look like this:

Ep = Ec + Ek;

m * g * H = n * m * g * H – Eк;

m * g * H – n * m * g * H = Eк;

m * g * H (1 – n) = Eк;

H = Eк / (m * g * (1 – n)) = 6 / (0.1 * 10 * (1 – 0.2)) = 6 / (0.1 * 10 * 0.8) = 6 / 0.8 = 7.5 meters.

Answer: The ball fell to a height of 7.5 meters.