Assuming a nearly frictionless ride, what can you say about a roller coaster’s potential and kinetic energy from the top to the bottom of a hill?

The mechanical energy of the roller coaster is sum of kinetic energy K and gravitational potential energy U:

where
is the kinetic energy
is the gravitational potential energy

Since the ride is frictionless, the total mechanical energy E is conserved during the ride. Therefore, at the top of the hill, the potential energy is maximum, because h (the height) is maximum, and this means the kinetic energy is minimum (because the sum of K and U is constant), so the velocity will be minimum. Viceversa, at the bottom of the hill, the potential energy will be minimum (because h is minimum), so the kinetic energy K will be maximum, and the velocity v of the roller coaster will be maximum.

Question was not very clear if there are specific answer choices tell me and i'll try to you more. hope this though2 laps is 1 min. we know that for each min she does two times the amount of laps. for 12 laps we can do 12/2 which does give us 12 laps in 6 mins not 16 mins.

Kinetic energy = 1/2mv^2. since the mass is constant we can form a ratio of it's kinetic energy at 60 divided by that at 30 to see ke(60)/ke(30) = +(60^2)/30^2). this may be said to 300% increase.