Confused on potential energy losses and regain.

[Lucyc2008]

Homework Statement

A 3.20kg rubber ball drops from a height of 4.80m to the ground a bounces back to a height of 2.70m
a) how much potential energy does the ball lose on the trip down?
b) how much energy does the ball regain on the trip back up?
c) what is the net loss of potential energy?

Homework Equations

Ep = mgh

The Attempt at a Solution

I got the question "correct" a) (3.20) x (9.80) x (4.80) = 151J b) (3.20) x (9.80) x (2.70) = 84.7J C) 151 - 84.7 = 66.3J but when studying for my test I found a completely different way of solving it by subtracting Ep2 (84.7) from Ep1 (151) so answer a) would be Ep lost = 67 instead. What is the correct way of solving this equation?. Attached is the second way of solving. The first way makes more sense in my mind because there is no height as the ball hits the ground so Ep = 0.

 

[kuruman]

Homework Statement

A 3.20kg rubber ball drops from a height of 4.80m to the ground a bounces back to a height of 2.70m
a) how much potential energy does the ball lose on the trip down?
b) how much energy does the ball regain on the trip back up?
c) what is the net loss of potential energy?

Homework Equations

Ep = mgh

The Attempt at a Solution

I got the question "correct" a) (3.20) x (9.80) x (4.80) = 151J b) (3.20) x (9.80) x (2.70) = 84.7J C) 151 - 84.7 = 66.3J but when studying for my test I found a completely different way of solving it by subtracting Ep2 (84.7) from Ep1 (151) so answer a) would be Ep lost = 67 instead. What is the correct way of solving this equation? The second way subtracting Ep2 from Ep1 makes more sense in my mind. Attached is the second way of solving.

In part (a) you are asked to find the loss of potential energy from the point of release to the point of contact with the floor. The answer to that is 151 J. The 67 J (or 66.3 J) is the answer to part (c), the net loss of potential energy which is also the loss of potential energy from 4.80 m to 2.70 m.