A roller coaster has a mass of 425 kg. It sits at the top of a hill with height 66 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.)

36.0 m/s

mgh = mv²/2

v=sqrt{2gh}

thank you ih

To determine the speed of the roller coaster when it reaches the bottom of the hill, we can use the principle of conservation of energy. At the top of the hill, the roller coaster has potential energy due to its height, and as it goes down, this potential energy is converted into kinetic energy.

The potential energy (PE) of an object depends on its mass (m), the acceleration due to gravity (g), and its height (h), and is given by the formula: PE = m * g * h.

In this case, the mass of the roller coaster is 425 kg, the acceleration due to gravity is approximately 9.8 m/s^2, and the height of the hill is 66 m. Plugging these values into the formula, we get: PE = 425 kg * 9.8 m/s^2 * 66 m = 276,852 J.

The potential energy at the top of the hill is converted into kinetic energy (KE) at the bottom, given by the formula: KE = 0.5 * m * v^2, where v is the velocity of the roller coaster at the bottom.

Since energy is conserved, we can equate the potential energy to the kinetic energy: PE = KE. Thus, we have: m * g * h = 0.5 * m * v^2.

Simplifying the equation, we find: v^2 = 2 * g * h. Plugging in the values for g (9.8 m/s^2) and h (66 m), we get: v^2 = 2 * 9.8 m/s^2 * 66 m = 1,293.6 m^2/s^2.

Finally, taking the square root of both sides, we find: v = √(1,293.6 m^2/s^2) = 35.96 m/s.

Therefore, the roller coaster will be traveling at approximately 35.96 m/s (or about 129.5 km/h) when it reaches the bottom of the hill.