A Ferris Wheel is built such that the height in feet above ground of a seat on the wheel at time t seconds can be modeled by h(t) = 53 + 50sin((pi/16)t - (pi/2)) The wheel makes one revolution every 32 seconds. The ride begins when t = 0. During the first 32 seconds of the ride, when will a person on the Ferris Wheel be 53 feet above the ground?

Answer:at time, t = 8 seconds and t = 24 seconds Ferris Wheel be 53 feet above the groundStep-by-step explanation:Data provided in the question:height in feet above ground of a seat on the wheel at time t seconds ismodeled ash(t) = [tex]53 + 50\sin((\frac{\pi t}{16} - \frac{\pi}{2})[/tex]now,at height 53 above the ground, we get the equation as:53 = [tex]53 + 50\sin(\frac{\pi t}{16} - \frac{\pi}{2})[/tex]or[tex]50\sin(\frac{\pi t}{16} - \frac{\pi}{2})[/tex] = 53 - 53or[tex]\sin(\frac{\pi t}{16} - \frac{\pi}{2})[/tex] = 0also,sin(0) = 0and, sin(π) = 0therefore,[tex](\frac{\pi t}{16} - \frac{\pi}{2})[/tex] = 0or[tex]\frac{\pi t}{16} = \frac{\pi}{2}[/tex]ort = 8 secondsand,[tex](\frac{\pi t}{16} - \frac{\pi}{2})[/tex] = πor[tex]\frac{\pi t}{16}= \pi + \frac{\pi}{2}[/tex] or[tex]\frac{\pi t}{16}= \frac{3\pi}{2}[/tex] ort = 24 secondsHence,the at time, t = 8 seconds and t = 24 seconds Ferris Wheel be 53 feet above the ground