We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $\triangle OBC$ is a $30-60-90$ triangle. That means $\angle BOC = 60^\circ$ and the ferris wheel has made $\frac{60}{360} = \frac{1}{6}$ of a revolution. Therefore, the time it takes to travel that much of a distance is $\frac{1}{6}\text{th}$ of a minute, or $10$ seconds. The answer is $\boxed{\mathrm{(D) \ } 10}$. Alternatively, we could also say that $\triangle ABC$ is congruent to $\triangle OBC$ by SAS, so $AC$ is 20, and $\triangle AOC$ is equilateral, and $\angle BOC = 60^\circ$
Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?
$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15$