# Homework Help: revolutions and trigonometry

Here's the question:

A Ferris wheel (d=50ft) makes 1 revolution/40 sec. If the center of the wheel is 30ft above the ground, how long after reaching the lowest point is a rider 50 ft above the ground?

I've tried a number of things, but I realize I'm probably just having trouble figuring out the arc length from the bottom of the wheel to the point where the rider's height from the bottom is 45. The y-value thing makes me think I need a sin-inverse. I thought it might be 45/50pi, but that's greater than 1. Basically, please help me find the arc length from the bottom of the wheel to where the y-value of the point is 45.

Thanks.

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Please provide us with what you have. – Jeel Shah Oct 6 '12 at 22:51
Well, I tried a couple things with the assumption the distance around the circle I was looking for was 45 ft, but that's obviously wrong. I tried (sorry for no LaTeX): 50pi ft/40 sec = (45/50pi ft)/x sec and other ratio things. I tried a number of sin-inverses: 45/50pi, 20/50pi (because r=25, 45-25=20) – Maddy Byahoo Oct 6 '12 at 22:56

For these type of problems it usually helps to draw diagrams.

Solving for $\theta$ $$\sin\theta=\frac{20}{25}$$ $$\theta=\sin^{-1}\left(\frac 4 5\right)$$ Therefore one must travel $\sin^{-1}\left(\frac 4 5\right)+90$ degrees to reach an altitude of 50m. Since we travel at a rate of $360^{\circ}/40 \rm \;sec=9^{\circ}/\rm \;sec$ it takes us $\frac{\left(\sin^{-1}\left(\frac 4 5\right)+90\right)}{9}\approx15.9 \rm \;sec$ to reach an altitude of 50m.

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+1 for an great diagram and clarity in answer – Maddy Byahoo Oct 7 '12 at 15:23