Question

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

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

Answer 1

Answer: (A), (D)

At an angle ${\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with the vertical.
${\overrightarrow{v}}_r=-30\hat{j},{\overrightarrow{v}}_m=\frac{10\sqrt{3}}{2}\hat{i}+\frac{10}{2}\hat{j}$
${\overrightarrow{v}}_{r/m}={\overrightarrow{v}}_r-{\overrightarrow{v}}_m=-5\sqrt{3}\hat{i}-35\hat{j}$
$\tan \theta =\frac{5\sqrt{3}}{35}=\frac{\sqrt{3}}{7}\Rightarrow \theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{7}\right)$ with vertical


Angle of $v_{r/m}$ with inclined plane is $60-\theta =\beta$
$\begin{array}{cc}\tan \beta =\tan (60-\theta )& =\frac{\tan {60}^{\circ }-\tan \theta }{1+\tan {60}^{\circ }\times \tan \theta }\\ & =\frac{\sqrt{3}-\sqrt{3}/7}{1+\sqrt{3}\times \sqrt{3}/7}=\frac{3\sqrt{3}}{5}\end{array}$
$\beta ={\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)$