2002 AMC 10B Problem 24

Put the center $O$ at height $20.$ The bottom $A$ is at height $0,$ and the rider reaches height $10,$ which is $10$ feet below the center. Draw a horizontal segment from the rider to the wheel’s vertical diameter. The resulting right triangle has a vertical leg of length $10$ and a hypotenuse (the radius) of length $20.$ That leg is half the hypotenuse, so the radius to the rider makes $60^\circ$ with the downward vertical. The wheel turns $360^\circ$ in $60$ seconds, so turning $60^\circ$ takes $\dfrac{60}{360}\cdot 60 = 10$ seconds. Thus, the correct answer is D.