Question

A man running on a horizontal road at $8\text{km/h}$ finds the rain falling vertically. He increases his speed to $12\text{km/h}$ and finds that the drops make an angle ${30}^{\circ }$ with the vertical. Find the speed of the rain with respect to the road (in $\text{km/hr}$).

• A. $2\sqrt{7}$
• B. $4\sqrt{7}$
• C. $5\sqrt{7}$
• D. $6\sqrt{7}$

Answer 1

The correct option is B $4\sqrt{7}$

${\overrightarrow{V}}_{\text{rain, road}}={\overrightarrow{V}}_{\text{rain, man}}+{\overrightarrow{V}}_{\text{man, road}}$ .....(i)

The two situations given in the problem may be respresented by the following figure.


${\overrightarrow{V}}_{\text{rain, road}}$ is same in magnitude and direction in both the figures.
Taking horizontal components in equation $\left(i\right)$ for figure $\left(a\right)$.
$V_{\text{rain, road}}\sin \alpha =8\text{km/h}........\left(ii\right)$
Here $\alpha$ is angle of ${\overrightarrow{V}}_{\text{rain, road}}$ with vertical.

Now consider figure $\left(b\right)$
$OA\perp V_{\text{rain, man}}$ as shown.
Taking components in equation $\left(i\right)$ along the line $OA$.
$V_{\text{rain, road}}\sin ({30}^{\circ }+\alpha )=12\cos {30}^{\circ }$ ....$\left(iii\right)$
From $\left(ii\right)$ and $\left(iii\right)$,
$\frac{\sin ({30}^{\circ }+\alpha )}{\sin \alpha }=\frac{12\times \sqrt{3}}{8\times 2}$
$\Rightarrow \frac{\sin {30}^{\circ }\cos \alpha +\cos {30}^{\circ }\sin \alpha }{\sin \alpha }=\frac{3\sqrt{3}}{4}$
$\Rightarrow \frac{1}{2}\cot \alpha +\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{4}$
i.e $\text{cot}\alpha =\frac{\sqrt{3}}{2}$ or $\alpha ={\cot }^{-1}\frac{\sqrt{3}}{2}$

From $\left(ii\right)$, $V_{rain,road}=\frac{8\text{km/h}}{\sin \alpha }$
$=4\sqrt{7}\text{km/h}$