# A soccer ball weighing 450 grams is thrown upward at a speed of 6 m / s.

**A soccer ball weighing 450 grams is thrown upward at a speed of 6 m / s. How high does it rise? Realize the problem statement. How many energy levels did you consider when solving the problem.**

Given:

m = 450 grams = 0.45 kilograms – the mass of a soccer ball that is thrown up;

g = 9.8 meters per second squared – gravitational acceleration (constant near the surface of the earth);

v0 = 6 meters per second – the initial speed at which the ball is thrown.

It is required to determine h (meter) – the height to which the ball will rise.

Since the condition of the problem is not specified, the air resistance is not taken into account when solving. Then, according to the law of conservation of energy, the kinetic energy of the ball will be equal to the potential energy of the ball at the maximum height:

Epot = Ekin;

m * g * h = m * v ^ 2/2;

g * h = v ^ 2/2;

h = v ^ 2 / (2 * g) = 6 ^ 2 / (2 * 9.8) = 36 / 19.6 = 1.8 meters (the result has been rounded to one decimal place).

Answer: the ball will rise to a height of 1.8 meters (when solving the problem, 2 energy levels are considered: during the ball throw at zero height and at the maximum height).