Solution to 2008 Problem 56

In order to fly north the plane should orient itself as if it were flying at an angle \theta west of north in still air, where
\begin{align*}\sin \theta = \left(\frac{30 \mbox{ km/h}}{200 \mbox{ km/h}} \right)\end{align*}
When it does this, the velocity in the northward direction is
\begin{align*}\sqrt{\left(200 \mbox{ km/h}\right)^2 - \left(30 \mbox{ km/h}\right)^2} = 10 \sqrt{391} \mbox{ km/h}\end{align*...
So, the time required to fly 500 \mbox{ km} north is
\begin{align*}\frac{500 \mbox{ km}}{10 \sqrt{391} \mbox{ km/h}} = \boxed{\frac{50}{\sqrt{391}} \mbox{ h}}\end{align*}
Therefore, answer (D) is correct.

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