Question

Rain pouring down at an angle $\alpha$ with the vertical has a speed of $5{\text{ms}}^{-1}$. A girl runs against the rain with a speed of $3{\text{ms}}^{-1}$ and sees that rain makes an angle $\beta$ with the vertical, then the relation between $\alpha$ and $\beta$ is

• A. $\tan \alpha =\tan \beta$
  
• B. $\tan \alpha =\cot \beta$
  
• C. $\tan \alpha =\frac{3+5\sin \beta }{5\cos \beta }$
  
• D. $\tan \beta =\frac{3+5\sin \alpha }{5\cos \alpha }$
  

Answer 1

The correct option is D $\tan \beta =\frac{3+5\sin \alpha }{5\cos \alpha }$


Given,
velocity of rain, $V_R=5{\text{ms}}^{-1}$
velocity of girl, $V_G=3{\text{ms}}^{-1}$


From figure, velocity of rain with respect to girl,
$\overrightarrow{V_{RG}}=\overrightarrow{V_R}-\overrightarrow{V_G}...\left(1\right)$

components of $\overrightarrow{V_{RG}}$
$\overrightarrow{V_{RG}}=V_{RG}(\sin \beta \hat{i}-\cos \beta \hat{j})$

Now, from figure components of $\overrightarrow{V_R}$
$\overrightarrow{V_R}=V_R(\sin \alpha \hat{i}-\cos \alpha \hat{j})$
$\Rightarrow \overrightarrow{V_R}=5(\sin \alpha \hat{i}-\cos \alpha \hat{j})$

components of $\overrightarrow{V_G}$
${\overrightarrow{V}}_G=-V_G\hat{i}=-3\hat{i}$

Substitute the above value in the equation $\left(1\right)$, we get
$V_{RG}(\sin \beta \hat{i}-\cos \beta \hat{j})=5(\sin \alpha \hat{i}-\cos \alpha \hat{j})+3\hat{i}$

$V_{RG}\sin \beta \hat{i}-V_{RG}\cos \beta \hat{j}=(5\sin \alpha +3)\hat{i}-5\cos \alpha \hat{j}$

Equating $\hat{i}$ and $\hat{j}$ on either side,
$V_{RG}\sin \beta =3+5\sin \alpha ...\left(2\right)$

$V_{RG}\cos \beta =5\cos \alpha ...\left(3\right)$

$\left(2\right)$ & $\left(3\right)$ gives,

$\tan \beta =\frac{3+5\sin \alpha }{5\cos \alpha }$

Hence, option (d) is correct answer.