91影视

The given function is$h\left(t\right)=125\sin \left(0.157t-\frac{\mathrm{蟺}}{2}\right)+125$wherethe heighth, in feet, of a seat on the wheel as a function of timet, wheret is measured in seconds. We have to find time for an individual on the Ferris wheel exactly 125above the ground.

To find the time, substitute 125 to the given function.

$$\begin{gathered} h\left(t\right)=125\sin \left(0.157t-\frac{\mathrm{蟺}}{2}\right)+125 \\ 125=125\sin \left(0.157t-\frac{\mathrm{蟺}}{2}\right)+125 \\ 0=125\sin \left(0.157t-\frac{\mathrm{蟺}}{2}\right) \\ 0=\sin \left(0.157t-\frac{\mathrm{蟺}}{2}\right) \\ {\sin }^{-1}\left(0\right)=\left(0.157t-\frac{\mathrm{蟺}}{2}\right) \\ \mathrm{Sine}\mathrm{function}\mathrm{is}\mathrm{equal}\mathrm{to}0\mathrm{when}\mathrm{the}\mathrm{angle}\mathrm{are}0\mathrm{and}\mathrm{蟺}. \\ \left(0.157\mathrm{t}-\frac{\mathrm{蟺}}{2}\right)=0\mathrm{or}\left(0.157\mathrm{t}-\frac{\mathrm{蟺}}{2}\right)=\mathrm{蟺} \\ 0.157\mathrm{t}=\frac{\mathrm{蟺}}{2}\mathrm{or}0.157\mathrm{t}=\mathrm{蟺}+\frac{\mathrm{蟺}}{2} \\ 0.157\mathrm{t}=1.57\mathrm{or}0.157\mathrm{t}=4.17 \\ \mathrm{t}=10\mathrm{or}\mathrm{t}=30 \end{gathered}$$

Thus, the individual on the Ferris wheel exactly 125 feet above on the ground is at time$10\mathrm{or}30\mathrm{seconds}.$

To find the time, substitute 250 to the given function.

$$\begin{gathered} h\left(t\right)=125\sin \left(0.157t-\frac{\mathrm{蟺}}{2}\right)+125 \\ 250=125\sin \left(0.157t-\frac{\mathrm{蟺}}{2}\right)+125 \\ 250-125=\sin \left(0.157t-\frac{\mathrm{蟺}}{2}\right) \\ \frac{125}{125}=\sin \left(0.157t-\frac{\mathrm{蟺}}{2}\right) \\ 1=\sin \left(0.157t-\frac{\mathrm{蟺}}{2}\right) \\ {\sin }^{-1}\left(1\right)=\left(0.157t-\frac{\mathrm{蟺}}{2}\right) \end{gathered}$$

Now, during the first 80seconds of the ride, the wheel will complete two revolutions. Thus, the interval will be $[0,4\mathrm{蟺}).$

$$\begin{gathered} \mathrm{Sine}\mathrm{function}\mathrm{is}\mathrm{equal}\mathrm{to}1\mathrm{when}\mathrm{the}\mathrm{angle}\mathrm{are}\frac{\mathrm{蟺}}{2}\mathrm{and}\frac{5\mathrm{蟺}}{2}. \\ \left(0.157\mathrm{t}-\frac{\mathrm{蟺}}{2}\right)=\frac{\mathrm{蟺}}{2}\mathrm{or}\left(0.157\mathrm{t}-\frac{\mathrm{蟺}}{2}\right)=\frac{5\mathrm{蟺}}{2} \\ 0.157\mathrm{t}=\frac{2\mathrm{蟺}}{2}\mathrm{or}0.157\mathrm{t}=\frac{5\mathrm{蟺}}{2}+\frac{\mathrm{蟺}}{2} \\ 0.157\mathrm{t}=\mathrm{蟺}\mathrm{or}0.157\mathrm{t}=3\mathrm{蟺} \\ 0.157\mathrm{t}=3.14\mathrm{or}0.157\mathrm{t}=9.42 \\ \mathrm{t}=20\mathrm{or}\mathrm{t}=60 \end{gathered}$$

Thus, an individual on the Ferris wheel exactly250feet above the ground at$20\mathrm{or}60\mathrm{seconds}.$

As we know that an individual on the Ferris wheel exactly 125feet above the ground at the time $10\mathrm{or}30\mathrm{seconds}.$

So, an individual on the Ferris Wheel more than 125feet above the ground at the time between intervals $10\mathrm{or}30\mathrm{seconds},10<\mathrm{t}<30.$

The interval will be$\left(10,30\right).$