A Ferris wheel has a radius of 10 m, and the bottom of the wheel passes 1 m above the ground. If the Ferris wheel makes one complete revolution every 18 s, find an equation that gives the height above the ground of a person on the Ferris wheel as a function of time

Answer:[tex]1+10(1-\cos (\frac{\pi t}{9}))[/tex]Step-by-step explanation:Givenradius of wheel [tex]r=10 m[/tex]Time period of Wheel [tex]T=18 s[/tex]and [tex]T\cdot \omega =2\pi [/tex] , where [tex]\omega =angular velocity of wheel[/tex][tex]\omega =\frac{2\pi }{18}[/tex]Let at any angle [tex]\theta [/tex]with vertical position of a point is given by [tex]x=r\sin \theta [/tex][tex]y=y_0+r(1-\cos \theta )[/tex]and [tex]\theta =\omega \times t[/tex]for velocity differentiate x and y to get [tex]v_x=r\cos \theta =r\cos (\omega t)[/tex][tex]v_y=0+r(\sin \theta )=r\sin (\omeag t)[/tex]Height at any time t is given by [tex]h=1+10(1-\cos \theta )=1+10(1-\cos (\frac{\pi t}{9}))[/tex]