Formula
The energy of a body's position in a gravitational field equals mass times free-fall acceleration times height: E = mgh.
ENT-level problem
A body of mass $m = 1$ kg slides down a slope of height $h = 5$ m without friction. Find the speed of the body at the bottom of the slope. The acceleration due to gravity is $g = 10$ m/s²; neglect air resistance.
There is no friction, so the kinetic energy at the bottom equals the potential energy at the top:
$$mgh = \frac{mv^2}{2}$$
The mass cancels, hence
$$v = \sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 5} = \sqrt{100} = 10 \text{ m/s}$$
The potential energy of a ball depends on its height h above the zero level. The higher the ball, the greater its potential energy. In the simulator the zero level is the lowest point of the trajectory.
The zero level is chosen arbitrarily — usually the ground. Rising to a height h increases the energy by mgh, lowering it decreases the energy; without friction the sum of kinetic and potential energy is conserved.
On the mass and the height above the chosen zero level: E = mgh. Free-fall acceleration on Earth is ≈ 9.81 m/s².
Zero if the ground is the zero level (h = 0). The level itself is a convention — only the height difference matters.
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