the blades of a windmill turn on an axis that is 35 feet above the ground. the blades are 10 feet long and…

the blades of a windmill turn on an axis that is 35 feet above the ground. the blades are 10 feet long and complete two rotations every minute. which of the following equations can be used to model h, the height in feet of the end of one blade, as a function of time, t, in seconds? assume that the blade is pointing to the right, parallel to the ground at t = 0 seconds, and that the windmill turns counterclockwise at a constant rate.\n\n$h=-10\\sin(\\frac{\\pi}{15}t)+35$\n\n$h=-10\\sin(\\pi t)+35$\n\n$h = 10\\sin(\\frac{\\pi}{15}t)+35$\n\n$h = 10\\sin(\\pi t)+35$

the blades of a windmill turn on an axis that is 35 feet above the ground. the blades are 10 feet long and complete two rotations every minute. which of the following equations can be used to model h, the height in feet of the end of one blade, as a function of time, t, in seconds? assume that the blade is pointing to the right, parallel to the ground at t = 0 seconds, and that the windmill turns counterclockwise at a constant rate.\n\n$h=-10\\sin(\\frac{\\pi}{15}t)+35$\n\n$h=-10\\sin(\\pi t)+35$\n\n$h = 10\\sin(\\frac{\\pi}{15}t)+35$\n\n$h = 10\\sin(\\pi t)+35$

Answer

Explanation:

Step1: Determine the amplitude

The length of the blade is 10 feet. This is the amplitude of the sinusoidal - function. Since the height of the axis is 35 feet above the ground, the vertical shift of the function is 35.

Step2: Calculate the angular frequency

The blades complete 2 rotations every minute. One rotation is (2\pi) radians. So in one minute (60 seconds), the blades rotate through an angle of (2\times2\pi = 4\pi) radians. The angular frequency (\omega) (in radians per second) is (\omega=\frac{4\pi}{60}=\frac{\pi}{15}) radians per second.

Step3: Determine the phase - shift and sign

At (t = 0), the blade is pointing to the right, parallel to the ground. We use a sine function. Since the windmill turns counter - clockwise, and we want the height to start at the mid - height (35 feet) and then go down first, the sign of the sine function is negative. The general form of a sinusoidal function for height (h) as a function of time (t) is (h = A\sin(\omega t+\varphi)+k), where (A) is the amplitude, (\omega) is the angular frequency, (\varphi) is the phase - shift, and (k) is the vertical shift. Here, (A=- 10), (\omega=\frac{\pi}{15}), (\varphi = 0), and (k = 35). So the function is (h=-10\sin(\frac{\pi}{15}t)+35).

Answer:

(h=-10\sin(\frac{\pi}{15}t)+35)