Step-by-step explanation
Given
From the question, we can see that the bottom of the ferris wheel is 30 feet about the ground and also while rotating, it can move to a height of 550 feet off the ground. In essence the actual height the ferris wheel can attain is
![h=550-30=520](https://img.qammunity.org/2023/formulas/mathematics/college/tjap0aewhcplb69j8l6zfgjpnra1vrn74t.png)
The amplitude then becomes half of the height which is
![A=(h)/(2)=(520)/(2)=260](https://img.qammunity.org/2023/formulas/mathematics/college/sge17352gxkcod89ms9yosd2gptjcsxsi8.png)
The vertical shift the Ferris wheel undergoes becomes the sum of the amplitude and its distance above the ground.
![D=260+30=290](https://img.qammunity.org/2023/formulas/mathematics/college/zysybf54a1tx317el206i0ij35xiyosvsc.png)
Since it takes the Ferris wheel 15 minutes to move from bottom to top, it will take it twice that to complete one revolution which will be its period.
![T=2*15mins=30](https://img.qammunity.org/2023/formulas/mathematics/college/3orq7ekoifbj02zr4mdtrzajy371phfla8.png)
Therefore, the frequency B, becomes;
![B=(2\pi)/(T)=(2\pi)/(30)=(\pi)/(15)](https://img.qammunity.org/2023/formulas/mathematics/college/uz0w2ejitty5uacv2xn6mewnqoxbm5zjt5.png)
We can then place in the above parameters to form the equation.
![y=Acos(B(t+C))+D\Rightarrow260cos((\pi)/(15)(t+C)+290](https://img.qammunity.org/2023/formulas/mathematics/college/6cwuidl339nvknbmtxpv4u4usu6h9kmss9.png)
The last missing parameter is the phase shift C. At time t =0, the function (y) has a position at 30. Therefore,
![\begin{gathered} 30=260cos((\pi)/(15)C)+290 \\ 260cos((\pi)/(15)C)=-290+30 \\ 260cos((\pi)/(15)C)=-260 \\ divide\text{ both sides by 260} \\ \frac{\begin{equation*}260cos((\pi)/(15)C)\end{equation*}}{260}=(-260)/(260) \\ cos((\pi)/(15)C)=-1 \\ (\pi)/(15)C=cos^(-1)(-1) \\ (\pi)/(15)C=\pi \\ C=(15\pi)/(\pi) \\ C=15 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/s91l1xx9vtplusyuqssr218qqaf8zj8oxd.png)
Therefore, the function y becomes
Answer:
![y=260cos((\pi)/(15)(t+15)+290](https://img.qammunity.org/2023/formulas/mathematics/college/g9pal51j6ay6y8fg5vh3tu57b9erklzo8q.png)