Question

The first hill of a roller coaster is 42.0 meters high. The roller coaster drops to a height of 4.2 meters above the ground at the bottom of this first hill.

If the mass of the roller coaster is 34,500 kg, what is the velocity at the bottom of the first hill? Ignore friction and assume the velocity at the top of the first hill is 0 m/s.

A. 1,614 m/s
B. 9.1 m/s
C. 20.3 m/s
D. 27.2 m/s

Answer 1

answer - d

set up equation

the velocity can be found through the change in kinetic energy

due to the conversation of energy, the change in kinetic energy is the same as the change in potential energy when the height is changed

ΔK =
`(mv^(2))/(2)`

ΔU = mgΔh

ΔK = ΔU

`(mv^(2))/(2)` = mgΔh

`v^(2)` = 2gΔh

v =
`√(2gh)`

values

g = 9.8
`(m)/(s^(2))`

Δh = 42 - 4.2 = 37.8 m

plug in values

v =
`√(2gh)`

v =
`√(2 * 9.8 * 37.8)`

v = 27.2
`(m)/(s)`

answer choice D