The energy that the ball had instantaneously prior to being dropped was Potential Energy (EP) only and gravitational potential energy is calculated as Ep = mgh = (2.80)(9.81)(4.50) = 123.606 Joules.
Because no outside energy was added and no work was done, the energy instantaneously prior to being dropped is equal to the energy when the mass is 3 m above the ground only at that point, the energy consists of two parts, gravitational Potential Energy (EP) and Kinetic Energy (EK). So, considering the point just prior to the drop as point 1 and the point 3 m above the ground as point 2, we can say: EP1 = EP2 + EK2 so EK2 = EP1 - EP2 and EP2 = (2.80)(9.81)(3.00) = 82.404 So EK2 = 123.606 - 82.404 = 41.2 Joules
They start as one stream, then get curvier, then empy into a delta. A delta is a ton of streams of water. It is one, then empties out into a ton of streams. ~Deceptiøn
