Question

A new ride being built at an amusement park includes a vertical drop of 71.6 meters. starting from rest, the ride vertically drops that distance before the track curves forward. if friction is neglected, what would be the speed of the roller coaster at the bottom of the drop? 13.2 m/s 23.6 m/s 26.5 m/s 37.5 m/s

Answer 1

The speed of the roller coaster at the bottom of the drop is approximately 26.5 m/s.

Step-by-step explanation:

To find the speed of the roller coaster at the bottom of the drop, we can use the principle of conservation of energy. At the top of the drop, the roller coaster has only gravitational potential energy, which is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. At the bottom of the drop, all of the gravitational potential energy is converted to kinetic energy, which is given by (1/2)mv^2, where v is the velocity. Equating the two expressions for energy, we have mgh = (1/2)mv^2. Canceling the mass on both sides and solving for v, we get v = sqrt(2gh). Substituting the given values of g = 9.8 m/s^2 and h = 71.6 m, we find v = sqrt(2 * 9.8 * 71.6) ≈ 26.5 m/s. Therefore, the speed of the roller coaster at the bottom of the drop is approximately 26.5 m/s.