8,000,000 kg of water are at the verge of dropping over niagara falls to the rocks 50.0 m below. what is the change in gravitational potential energy as the water splashes on the rocks below? ( calculate gpe final and gpe initial and then find the difference)

i will very much appreciate it you!

The Change in the Gravitational Potential Energy as the water splashes out on the rocks below is 3.92 * 10^9 J.

Explanation:

If an object possess energy  because of its position in a gravitational field is called as Gravitational Potential Energy.The formula for the Gravitational Potential Energy is given by,

                                        GPE = mgh

where m represents the mass in kilograms,

            g represents acceleration due to gravity (9.8 m/s^2 in Earth)

            h is the height in meters.

mass = 8,000,000 kg, g = 9.8 m/s^2,   initial height = 0

Initial GPE = 0

Final height = 50.0 m.

Final GPE = 8,000,000 * 9.8 * 50.0 = 3.92 * 10^9 J.

Change in Gravitational Potential Energy = Final GPE - Initial GPE

                                                                        = (3.92 * 10^9) - 0

                                                                        = 3.92 *10^9 J.

they should be similar or the same