A body weighing 0.5 kg. thrown upward at a speed of 30 m / s. What is the work done by the force of gravity

A body weighing 0.5 kg. thrown upward at a speed of 30 m / s. What is the work done by the force of gravity: when lifting to the maximum height; when it falls to its original level?

In order to find the work of gravity, we write down its formula:
A = Fs = mgh.

Now you need to calculate the height to which the body was lifted.
Let’s express it from the distance formula:

h = ut = (u + u0) / 2 * (u – u0) / g = u ^ 2 – u0 ^ 2 / 2g.

In the 1st case, the final speed is zero, we get:

u = 0;

h1 = u0 ^ 2 / 2g = 30 ^ 2 / (2 * 10) = 900/20 = 45 m.

In the second case, the initial velocity is zero, we get:

u0 = 0;

h2 = u ^ 2 / 2g = 30 ^ 2 / (2 * 10) = 900/20 = 45 m.

As a result, the total height is:
h = h1 + h2 = 45 + 45 = 90 m.

Let’s substitute it in the formula for the work of gravity:

A = Fs = mgh = 0.5 * 10 * 90 = 450 J.

Answer: 450 J.