A roller coaster has a mass of 275 kg. It sits at the top of a hill with height 85 m. If it drops from this hill, how fast is it going when it reaches the bottom? (Assume there is no air resistance or friction.)A.40.8 m/sB.73.4 m/sC.29.9 m/sD.18.0 m/s

To solve this problem, we can use the principle of conservation of energy. This principle states that the total energy in a closed system remains constant. In this case, the roller coaster's total energy is the sum of its potential energy (due to its height) and its kinetic energy (due to its motion).

At the top of the hill, the roller coaster has maximum potential energy and zero kinetic energy. At the bottom of the hill, it has maximum kinetic energy and zero potential energy. Therefore, we can set the potential energy at the top equal to the kinetic energy at the bottom and solve for the speed at the bottom.

The potential energy (PE) at the top of the hill is given by the formula PE = mgh, where m is the mass (275 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (85 m).

PE = 275 kg * 9.8 m/s^2 * 85 m = 229325 Joules

The kinetic energy (KE) at the bottom of the hill is given by the formula KE = 1/2 mv^2, where m is the mass (275 kg) and v is the speed. We can set this equal to the potential energy and solve for v:

229325 Joules = 1/2 * 275 kg * v^2

Rearranging the equation gives:

v^2 = 2 * 229325 Joules / 275 kg = 1670.18

Taking the square root of both sides gives:

v = sqrt(1670.18) = 40.87 m/s

So, the speed of the roller coaster at the bottom of the hill is approximately 40.87 m/s. The closest answer to this is A. 40.8 m/s.