A roller coaster starts from rest at the top of a 20.0 m hill, with friction assumed negligible. What is its final speed at the bottom, taking g = 9.80 m/s^2?

1. With friction negligible, the loss in gravitational potential energy equals the gain in kinetic energy, mg times the height h equals one half m v squared. 2. The mass cancels, leaving v equal to the square root of 2 times g times h. 3. Substituting the values, v = the square root of 2 times 9.80 m/s^2 times 20.0 m, which is the square root of 392 m^2/s^2. 4. Taking that square root gives v = 19.8 m/s. 5. 39.2 m/s comes from forgetting to take the square root, and 14.0 m/s and 9.80 m/s come from arithmetic slips inside the square root or from using g directly as if it were the speed. _Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.3 Gravitational Potential Energy_