A200 kg roller coaster car is located at the top of a hill where the height is 30 m. what is the velocity when it reaches the bottom of the hill?

Assuming there is not loss of energy due to frinction (which is an idealization), you can use the conservation of energy law in the form:

Mechanical energy at top = mechanical energy at bottom.

Mechanical energy at top = potential energy = m * g * h

Mechanical energy at bottom = kinetic energy = m * v^2 / 2

=> m *v^2 / 2 = m * g * h

=> v^2 / 2 = g * h

=> v ^2 = 2gh

=> v = √(2gh) = √(2 * 9.8 m/s^2 *30m) = 24.25 m/s

24.25 m/s

Particle exchange energy though elastic collision , hope this .