Passengers in the jet transport A flying with its nose pointed east at an air speed of $800 k m h^{- 1}$ observe a second jet plane B that passes under the transport in horizontal flight. Although the nose of B is pointed in the $45^{\circ}$ northeast direction, plane B appears to the passengers in A to be moving away from the transport at the $60^{\circ}$ angle as shown. If the true air speed of B is 100n kph, find the integer closest to n. (Air speed is velocity with respect to air, Assume a wind towards south west at 100kph)

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answer is 7.

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Detailed Solution

$\begin{array}{l} V_{A} = 800 \hat{i} \\ V_{B} = \frac{V_{B}}{\sqrt{2}} \hat{i} + \frac{V_{B}^{A}}{\sqrt{2}} \hat{j} \\ \frac{V_{B}}{A} = V_{B} - V_{A} = - (800 - \frac{V_{B}}{\sqrt{2}}) \hat{i} + \frac{V_{B}}{\sqrt{2}} \hat{j} \\ \tan 60^{\circ} = \frac{\frac{V_{B}}{\sqrt{2}}}{800 - \frac{V_{B}}{\sqrt{2}}} \\ \Rightarrow \sqrt{3} \times 800 = \frac{V_{B}}{\sqrt{2}} (1 + \sqrt{3}) \\ V_{B} = \frac{800 \sqrt{3} \times \sqrt{2}}{1 + \sqrt{3}} \\ = 717 km / h \end{array}$