Roller Coaster Physics: Velocity at Bottom of Hill

Answer: 22.4 m/s

Explanation

Using the conservation of energy principle, the total mechanical energy of the roller coaster car (a closed system) remains constant. At the top of the hill, the car has only potential energy (PE = mgh). At the bottom, it has only kinetic energy (KE = ½mv²). Equating the two:

mgh = ½mv² → v = √(2gh)

Given potential energy PE = 20,000 J, mass m = 250 kg, and g = 9.8 m/s², we can find the height h = PE/(mg) = 20,000/(250×9.8) ≈ 8.16 m. Then v = √(2×9.8×8.16) ≈ 22.4 m/s.

Therefore, the velocity at the bottom of the hill is 22.4 m/s.