Question

A new ride being built at an amusement park includes a vertical drop of 126.5 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?
17.60 m/s
24.90 m/s
49.79 m/s
70.42 m/s

Answer 1

Answer:

Speed at the bottom of the roller coaster = 49.79 m/s

Explanation:

A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically drops that distance before the track curves forward.

We have to find the speed at the bottom.

Here the gravitational energy fully converts to kinetic energy, so we equate it.

Gravitational energy = $m\times g\times h$

Kinetic energy = $0.5 \times m\times v^{2}$

$m\times g\times h$ = $0.5 \times m\times v^{2}$

$9.8\times 126.5 = 0.5\times v^{2}$

$v^{2}$ = 2479.4

Velocity, v = 49.79 m/s