Question

A stationary person observes that rain is falling vertically down at $30km/hr$. A cyclist is moving up on an inclined plane making an angle ${30}^{\circ }$with horizontal at $10km/hr$. In what direction should the cyclist hold his umbrella to prevent himself from rain?

• A. At an angle ${\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$ with the inclined plane
• B. At an angle ${\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with the inclined plane
• C. At an angle ${\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$ with the horizontal
• D. At an angle ${\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with the vertical

Answer 1

Answer: (A), (D)

At an angle ${\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$ with the inclined plane
At an angle ${\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with the vertical

${\overrightarrow{v}}_m=-30\hat{j},{\overrightarrow{v}}_m=\frac{10\sqrt{3}}{2}\hat{i}+\frac{10}{2}\hat{j}$
$\tan \theta =\frac{5\sqrt{3}}{35}=\frac{\sqrt{3}}{7}\Rightarrow \theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with vertical
Angle of $v_{r/m}$ with inclined plane is $60-\theta =\beta$
$\tan \beta =\tan ({60}^0-\theta )=\frac{\tan {60}^{\circ }-\tan \theta }{1+\tan {60}^{\circ }\cdot \tan \theta }$
$=\frac{\sqrt{3}-\sqrt{3}/7}{1+\sqrt{3}\cdot \sqrt{3}/7}=\frac{3\sqrt{3}}{5}$
$\Rightarrow \beta ={\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$.