A stationary person observes that rain is falling vertically down at 30 kmh^-1. A cyclist is moving up on an inclined plane making an angle 30^o with horizontal at 10 kmh^-1. In which direction should the cyclist hold his umbrella to prevent himself from the rain?

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At an angle $\tan^{- 1} (\frac{3 \sqrt{3}}{5})$ with inclined plane

At an angle $\tan^{- 1} (\frac{3 \sqrt{3}}{5})$ with horizontal

At an angle $\tan^{- 1} (\frac{\sqrt{3}}{7})$ with inclined plane

At an angle $\tan^{- 1} (\frac{\sqrt{3}}{7})$ with vertical

answer is A, D.

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Detailed Solution

${\vec{v}}_{r} = - 30 \hat{j}, {\vec{v}}_{m} = \frac{10 \sqrt{3}}{2} \hat{i} + \frac{10}{2} \hat{j}$

$\tan θ = \frac{5 \sqrt{3}}{35} = \frac{\sqrt{3}}{7} \Rightarrow θ = \tan^{- 1} (\frac{\sqrt{3}}{7})$ with vertical

Angle of v_r/m with inclined plane is

$\begin{array}{l} 60^{\circ} - θ = β \\ \tan β = \tan (60^{\circ} - θ) \\ = \frac{\tan 60^{\circ} - \tan θ}{1 + \tan 60^{\circ} \tan θ} \\ = \frac{\sqrt{3} - \sqrt{3} / 7}{1 + \sqrt{3} \cdot \sqrt{3} / 7} = \frac{3 \sqrt{3}}{5} \end{array}$

$\Rightarrow β = \tan^{- 1} (\frac{3 \sqrt{3}}{5})$

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