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?

1) 13.2 m/s
2) 23.6 m/s
3) 26.5 m/s
4) 37.5 m/s

Answer 1

The speed of the roller coaster at the bottom of a 71.6-meter vertical drop is calculated using the conservation of energy principle. Given that no energy is lost to friction, the potential energy at the top is converted into kinetic energy at the bottom. The correct speed at the bottom is approximately 37.5 m/s.

Step-by-step explanation:

To find out the speed of the roller coaster at the bottom of the 71.6-meter vertical drop, you can use the principle of conservation of energy. Assuming no energy is lost to friction, the potential energy at the top of the drop will be converted into kinetic energy at the bottom. The potential energy (PE) at the top given by PE = m * g * h, where m is the mass, g is the acceleration due to gravity (9.8 m/s2), and h is the height. This will be equal to the kinetic energy (KE) at the bottom given by KE = (1/2) * m * v2, where v is the speed we want to find.

So, setting the potential energy equal to the kinetic energy and assuming mass cancels out, we get:

g * h = (1/2) * v2

Solving for v gives us:

v = √(2 * g * h)

Plugging in our values gives:

v = √(2 * 9.8 m/s2 * 71.6 m)

v = √(2 * 9.8 * 71.6)

v = √(1403.36)

v = 37.46 m/s

Therefore, the roller coaster would be traveling at a speed of approximately 37.5 m/s at the bottom of the drop, making option 4) 37.5 m/s the correct choice.