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.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.