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 $10km/hr$. At what angle with the inclined plane the cyclist should hold his umbrella to protect himself from rain?

• A. ${\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$
• B. ${\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$
• C. ${\tan }^{-1}\left(\frac{3}{5}\right)$
• D. ${\tan }^{-1}\left(\frac{3\sqrt{3}}{7}\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$


$\overrightarrow{v_r}=-30\hat{j}$​,
$\overrightarrow{v_m}=10\cos {30}^{\circ }\hat{i}+10\sin {30}^{\circ }\hat{j}=5\sqrt{3}\hat{i}+5\hat{j}$​
$\overrightarrow{v_{r,m}}=\overrightarrow{v_r}-\overrightarrow{v_m}=-5\sqrt{3}\hat{i}-35\hat{j}$

If $\theta$ is the angle made by $\overrightarrow{v_{r,m}}$ with vertical,

$\tan \theta =\frac{v_{r,m_x}}{v_{r,m_y}}=\frac{5\sqrt{3}}{35}=\frac{\sqrt{3}}{7}$​​
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with vertical

Angle of ${\overrightarrow{v}}_{r,m}$​ with inclined plane is ${60}^{\circ }-\theta =\beta$

Then, $\tan \beta =\tan ({60}^{\circ }-\theta )=\frac{\tan {60}^{\circ }-\tan \theta }{1+\tan {60}^{\circ }\tan \theta }$
$=\frac{\sqrt{3}-\sqrt{3}/7}{1+\sqrt{3}.\sqrt{3}/7}$
$=\frac{3\sqrt{3}}{5}$
$⟹\beta ={\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$