Question

A theme park wants to build a roller coaster with a maximum velocity of 33.5 M/S (75 mph) how high must the first hill be in order to reach the maximum velocity?

A) 57.3 m
B) 287 m
C) 164.2 m
D) 99.1 m

Answer 1

Approximately
`57.2\; {\rm m}` (assuming no energy loss, and that the gravitational field strength is
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

Assume that there is no energy loss (e.g., friction or drag on the roller coaster). Also assume that there is no external energy input to the rollercoaster during the descent. The entirety of the kinetic energy (
`\text{KE}`) of this rollercoaster would need to be converted from gravitational potential energy (
`\text{GPE}`.)

Let
`m` denote the mass of this rollercoaster.

• At a height of
  `h`, the
  `\text{GPE}` of this rollercoaster would be
  `m\, g\, h`.
• When the rollercoaster is moving at a speed of
  `v`, the
  `\text{KE}` of this rollercoaster would be
  `(1/2)\, m\, v^(2)`.

The
`\text{KE}` of this rollercoaster is maximized when the speed of this rollercoaster is maximized. Let
`v_\text{max}` denote the maximum speed that this rollercoaster needs to reach. The maximum
`\text{KE}\!` that this rollercoaster need to reach would be:

`\begin{aligned} (1)/(2)\, m\, {v_\text{max}}^(2)\end{aligned}`.

Let
`h_(0)` denote the initial height of this rollercoaster. The initial
`\text{GPE}` of this rollercoaster would be
`m\, g\, h_(0)`. Since the entirety of the
`\text{KE}` of this rollercoaster needs to be converted from this
`\text{GPE}\!`,
`m\, g\, h_(0)\!` should be greater than or equal to
`(1/2)\, m\, {v_\text{max}}^(2)`:

`\begin{aligned} m\, g\, h_(0) &= (1)/(2)\, m\, {v_\text{max}}^(2)\end{aligned}`.

Solve this equation for
`h_(0)`:

`\begin{aligned}g\, h &= (1)/(2)\, {v_\text{max}}^(2)\end{aligned}`.

`\begin{aligned} h &= \frac{{v_\text{max}}^(2)}{2\, g}\end{aligned}`.

Given that
`v_\text{max} = 33.5\; {\rm m\cdot s^(-1)}` and assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`:

`\begin{aligned} h &= \frac{{v_\text{max}}^(2)}{2\, g} \\ &= \frac{(33.5\; {\rm m\cdot s^(-1)})^(2)}{2 * (9.81\; {\rm m\cdot s^(-2)})} \\ &\approx 57.2\; {\rm m} \end{aligned}`.

(Note that all values in this equation are in standard units.)

Thus, the height of the track should be at least (approximately)
`57.2\; {\rm m}`.