The law of conservation of energy states that the total energy in an isolated system is constant. Energy is not created or destroyed; it changes form. For our purposes, total energy is the sum of kinetic and potential energies.

When a ball is thrown upwards, it is gaining potential energy and losing kinetic energy joule for joule. When it starts falling back down, the trade occurs in reverse. Mathematically,

`Delta E_"k" + Delta E_"g" = 0`.

When we substitute those energy changes, we can rearrange to get

`v_2^2 - v_1^2 = -2gDelta h`.

This looks very similar to the eighth and final equation that we derived in the section on the equations of motion of the first unit.

If we were instead considering a situation where kinetic energy was converted to elastic energy (a ball hits a spring and compresses it), our equation would be `Delta E_"k" + Delta E_"e" = 0`.

A 2.5 kg block is dropped from 7.5 m above the floor. What is its speed as it hits the floor?

We can used the equation that I just mentioned:

`v_2^2 - v_1^2 = -2gDelta h`.

Solving for `v_2`, we have

`v_2 = sqrt(v_1^2-2gDelta h) = sqrt(0-2(9.80\ "N/kg")(- 7.5\ "m")) ~~ 12.124`,

therefore the final speed is 12 m/s.