To solve this problem, we need to model the height [tex]\( h \)[/tex] of the end of one windmill blade as a function of time [tex]\( t \)[/tex] in seconds.
### Step 1: Determine the Parameters
- The axis of rotation for the blades is 35 feet above the ground.
- The length of each blade is 10 feet.
- The windmill completes two rotations every minute.
### Step 2: Calculate the Rotation Period
Since the windmill completes two rotations every minute, each rotation takes:
[tex]\[
\text{One rotation period} = \frac{60 \text{ seconds}}{2 \text{ rotations}} = 30 \text{ seconds}
\][/tex]
### Step 3: Define the Amplitude
The amplitude of the height variation is equal to the length of the blade, which is:
[tex]\[
\text{Amplitude} = 10 \text{ feet}
\][/tex]
### Step 4: Define the Median Height
The median height where the blade is parallel to the ground (at [tex]\( t = 0 \)[/tex] seconds) is:
[tex]\[
\text{Median height} = 35 \text{ feet}
\][/tex]
### Step 5: Determine the Angular Frequency
The angular frequency [tex]\( \omega \)[/tex] for the sinusoidal function can be determined by using the period of rotation. The period [tex]\( T \)[/tex] is 30 seconds. The angular frequency is given by:
[tex]\[
\omega = \frac{2\pi}{T}
\][/tex]
Substituting the period [tex]\( T = 30 \)[/tex] seconds:
[tex]\[
\omega = \frac{2\pi}{30} = \frac{\pi}{15}\ \text{radians per second}
\][/tex]
### Step 6: Formulate the Sinusoidal Equation
We need to decide between [tex]\(\sin\)[/tex] and [tex]\(\cos\)[/tex]. At [tex]\( t = 0 \)[/tex], the blade is parallel to the ground, indicating we can use either function with appropriate phase adjustments. However, a [tex]\(\sin\)[/tex]-based function starting at the median height simplifies without needing phase shifts.
Given that the amplitude is 10 feet, the angular frequency is [tex]\( \frac{\pi}{15} \)[/tex], and the median height is 35 feet, the height [tex]\( h \)[/tex] can be modeled by the sinusoidal equation:
[tex]\[
h(t) = 10 \sin \left(\frac{\pi}{15} t\right) + 35
\][/tex]
### Step 7: Select the Correct Equation
Comparing this equation with the provided options:
- [tex]\( h = -10 \sin \left(\frac{\pi}{15} t\right) + 35 \)[/tex]
- [tex]\( h = -10 \sin (\pi t) + 35 \)[/tex]
- [tex]\( h = 10 \sin \left(\frac{\pi}{15} t\right) + 35 \)[/tex]
- [tex]\( h = 10 \sin (\pi t) + 35 \)[/tex]
The correct equation is:
[tex]\[
h = 10 \sin \left(\frac{\pi}{15} t\right) + 35
\][/tex]
So, the correct choice is:
[tex]\[
\boxed{h=10 \sin \left(\frac{\pi}{15} t\right) + 35}
\][/tex]
