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?

Answer 1

Answer:

37.5 m/s

Explanation:

Using the law of conservation of energy, potential energy equals kinetic energy hence

$mgh=0.5mv^{2}$

Therefore

$v=\sqrt{2gh}$

where g is the acceleration due to gravity, m is the mass of the object, h is the height and v is the speed of the roller coaster

Taking g as 9.81 then

$v=\sqrt{2\times 9.8\times 71.6}=37.48055496 m/s\approx 37.5 m/s$