# Height above ground at time t, of ferris wheel

• Apr 7th 2011, 01:28 PM
sydewayzlocc
Height above ground at time t, of ferris wheel
You have been riding the London Ferris Wheel for 17 minutes and 30 seconds, and the diameter of the wheel is 450 feet. What is your height above the ground at this time? Any help would be appreciated i am not sure how to go about doing this problem.
• Apr 7th 2011, 01:41 PM
e^(i*pi)
Not enough information, we need to know the period of frequency of the wheel
• Apr 7th 2011, 01:45 PM
sydewayzlocc
My mistake. The Ferris Wheel makes one complete revolution every 30 minutes
• Apr 7th 2011, 02:07 PM
e^(i*pi)
You can find the angle you're at relative to the bottom of the wheel by $\dfrac{\alpha}{360} = \dfrac{17.5}{30}$
Using the middle of the wheel as a point we can say that the sum of the angles around it is 360 degrees.

$\theta = 360 - \alpha - 90$

You can then construct a right angled triangle using $\theta$ as your angle. At your point you are $h + r$ feet above the ground where $h = r \sin \theta$

edit: scanned

P is the point you're at
• Apr 8th 2011, 07:01 AM
Soroban
Hello, sydewayzlocc!

Quote:

You have been riding the London Ferris Wheel for 17 minutes and 30 seconds,
and the diameter of the wheel is 450 feet. It makes one revolution every 30 minutes.
What is your height above the ground at this time?

Code:

             * * *         A*----*B  *         *  \  |      *       *    \  |      *           225\@|       *      \|        *       *        *O      *       *        |        *               |       *      |225    *         *      |      *           *    |    *             * * *               C

The wheel turns $360^o$ every 30 minutes: $12^o$ per minute.

In 17.5 minutes, it has turned $210^o\!:\;\theta \:=\:\angle AOB = 30^o$

Hence: . $BO \:=\:225\cos30^o \:=\:\dfrac{225\sqrt{3}}{2}$

Therefore: . $BO \;=\;225+\dfrac{225\sqrt{3}}{2} \;=\;\dfrac{225}{2}(2 + \sqrt{3})\text{ ft.}$