A 250.0 kg roller coaster car has 20,000.0 J of potential energy at the top of the hill. Neglecting frictional losses, what is the velocity of the car at the bottom of the hill?

Answer 1

The kinetic energy (1/2) M V^2 will equal the lost potential energy (20,000 J) at the bottom of the hill. This assumes that it had nearly zero kinetic energy at the top of the hill.

You know M, so solve for V.

V = sqrt(2*20,000/M)