Question

When a man moves down an inclined plane with a constant speed $5\text{m/s}$ which makes an angle of ${37}^{\circ }$ with the horizontal, he finds that the rain is falling vertically downwards. When he moves up the same inclined plane with the same speed, he finds that the rain makes an angle $\theta ={\tan }^{-1}\left(\frac{7}{8}\right)$ with the horizontal. The speed of the rain with respect to the ground is

• A. $\sqrt{116}\text{m/s}$
• B. $\sqrt{32}\text{m/s}$
• C. $\sqrt{10}\text{m/s}$
• D. $5\text{m/s}$

Answer 1

The correct option is B $\sqrt{32}\text{m/s}$

Let the velocity of rain w.r.t ground be ${\overrightarrow{V}}_r=a\hat{i}+b\hat{j}$
First case:


Velocity of man w.r.t ground, ${\overrightarrow{V}}_m=-4\hat{i}-3\hat{j}$
Velocity of rain w.r.t man,
$\Rightarrow {\overrightarrow{V}}_{rm}={\overrightarrow{V}}_r-{\overrightarrow{V}}_m=a\hat{i}+b\hat{j}-(-4\hat{i}-3\hat{j})$
$\Rightarrow {\overrightarrow{V}}_{rm}=(a+4)\hat{i}+(b+3)\hat{j}...\left(1\right)$
As rain seems to be falling vertically to the man
$\theta =-{90}^{\circ }$ [with x-axis]
$\Rightarrow \tan (-{90}^{\circ })\to -\infty$
$[\tan \theta =\frac{\text{Y-component}}{\text{X-component}}]$
From Eq.$\left(1\right)$,
$\frac{b+3}{a+4}\to -\infty$
$\Rightarrow a+4=0$
Or, $a=-4...\left(2\right)$

Second case-


Velocity of man w.r.t ground,
${\overrightarrow{V}}_m=4\hat{i}+3\hat{j}$
$\Rightarrow {\overrightarrow{V}}_{rm}={\overrightarrow{V}}_r-{\overrightarrow{V}}_m=a\hat{i}+b\hat{j}-(4\hat{i}+3\hat{j})$
$\Rightarrow {\overrightarrow{V}}_{rm}=(a-4)\hat{i}+(b-3)\hat{j}$
Substituting From Eq.$\left(2\right)$,
$\Rightarrow {\overrightarrow{V}}_{rm}=-8\hat{i}+(b-3)\hat{j}...\left(3\right)$
As rain seems to be falling at angle of $\theta ={\tan }^{-1}\left(\frac{7}{8}\right)$ with the horizontal (x-axis) w.r.t the man
$\tan \theta =\frac{7}{8}$
$\Rightarrow \frac{b-3}{-8}=\frac{7}{8}$
$[\because \tan \theta =\frac{\text{Y-component}}{\text{X-component}}]$
$\Rightarrow b-3=-7$
$\therefore b=-4$
Hence, velocity of rain w.r.t ground is
${\overrightarrow{V}}_r=a\hat{i}+b\hat{j}$
$=-4\hat{i}-4\hat{j}$
Now, its magnitude is
$|{\overrightarrow{V}}_r|=\sqrt{(-4{)}^2+(-4{)}^2}$
$\therefore |{\overrightarrow{V}}_r|=\sqrt{32}\text{m/s}$
The speed of rain w.r.t ground is $\sqrt{32}\text{m/s}$