How to write an equation using time as the independent variable representing trey's height above the ground as he rides the Ferris wheel

Question

1. A radius of 45 feet with a building to the left 
2. The height from the ground at the highest point on the wheel is 95 ft 
3. The distance from the building at the closest point is 12 feet 
4. The wheel rotates counterclockwise taking 120 seconds to complete 1 rotation 
5. Trey rides on the Ferris wheel starting to the far right of the center at point A

Answer 1

My teaching style is more focused on trying to get you to explore, rather than supply answers.

Have a look at this animation I made (paste this link into your browser):

https://drive.google.com/file/d/1K6cp1vpYXznE9sCB8TjiJyZjnC05eUNJ/view?usp=sharing

Sorry that Trey is upside down half the time (still working on this stuff :)).

It is critical that you pause the video and write down your work. Once you have found your function, you can check it at desmos.com

Answer 2

for trig sine, general equation is y=Asin(B(t-C))+D

The point A is given as center of wheel, so trey is 50 feet up at time zero. (95-90)+45

D=50

B is the rate (radians/sec) and is given as 2π/120

since trey is at same height as wheel center at t=0 there is no phase shift, that is C=0

at t=30 trey is 45 feet above wheel center meaning A=45

check, at t=90, trey should be at a height of 5.

For cosine y=Ecos(F(t-G))+H is general equation, the center line is 50 feet up, H=50.

recall y is height of trey..and at t=0 y=50

more:,,,,F is given as 2π/120, and E is the height above center line, E=45

have at t=0, 50=45*cos(-2Gπ/120)+50

means cos(-2Gπ/120)=0, or- 2Gπ/120=-π/2,

or G=30

y=45cos(2π/120[t-30])+50

check, at t=90 trey should be at 5 feet.