Question

Josie is sitting on a Ferris wheel. She is exactly 29 feet from the center and is at the 3 o'clock position when the Ferris wheel begins moving.Suppose Josie has rotated 3.5 radians around the Ferris wheel (starting at the 3 o'clock position). How high is Josie above the horizontal diameter (in feet)? _____feet   Suppose Josie has traveled 237 feet from the 3 o'clock position. How high is Josie above the horizontal diameter (in feet)?____ feet   Define a function f that determines Josie's vertical distance above the horizontal diameter (in feet) as a function of the arc length (in feet), s, swept out by the Ferris wheel as it moves counter-clockwise from the 3 o'clock position.

Answer 1

The radius of ferris wheel is r = 29 feet.

The equation for the vertical distance is,

`y=r\sin \theta`

For rotation of 3.5 radians,

`\begin{gathered} \theta=3.5*(180)/(\pi) \\ =200.53^(\circ) \end{gathered}`

Determine the vertical distance of Josie.

`\begin{gathered} y=29\cdot\sin 200.53 \\ =29\cdot(-0.3508) \\ =-10.1732 \end{gathered}`

Josie is 10.1732 feet below the horizontal.

Part B:

The distance travelled by Josie is s = 237 feet.

Determine the rotation for distance travelled by Josie.

`\begin{gathered} \theta=(s)/(r) \\ =(237)/(29) \\ =468.244^(\circ) \end{gathered}`

Determine the vertical distance of Josie.

`\begin{gathered} y=29\cdot\sin (468.244) \\ =29\cdot0.949 \\ =27.54 \end{gathered}`

Josie is 27.544 feet above the horizontal.

Part C:

The function for the vertical distance above the horizontal is,

`f(s)=29\sin ((s)/(29))`

Here angle is in radians.