Question

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

• A. $\sqrt{116}m{s}^{-1}$
• B. $\sqrt{32}m{s}^{-1}$
• C. $5m{s}^{-1}$
• D. $\sqrt{73}m{s}^{-1}$

Answer 1

Answer: (B)

$\sqrt{32}m{s}^{-1}$

Let velocity of rain is
${\overrightarrow{V}}_R=a\hat{i}-b\hat{j}$
In first case, ${\overrightarrow{V}}_M=(-4\hat{i}-3\hat{j})$
$\therefore {\overrightarrow{V}}_{RM}={\overrightarrow{V}}_R-{\overrightarrow{V}}_M=(a+4)\hat{i}+(-b+3)\hat{j}$
It appears vertical
$\therefore a+4=0$ or $a=-4$
In second case, ${\overrightarrow{V}}_M=(4\hat{i}+3\hat{j})$
$\therefore {\overrightarrow{V}}_{RM}={\overrightarrow{V}}_M=(a-4)\hat{i}+(-b-3\hat{j})$
$=-8\hat{i}+(-b-3\hat{j})$
It appears at $\theta =ta{n}^{-1}\left(\frac{7}{8}\right)$
$\therefore -b-3=-7$
$\therefore b=4$
and speed of rain $=\sqrt{a^2+b^2}=\sqrt{32}m/s$