When a man moves at a speed of 5 m/s down an inclined plane which makes angle of $37°$ with the horizontal, he finds the rain falling vertically downward. When he moves up the same inclined plane at the same speed, he finds the rain makes angle $\theta ={\tan }^{-1}(7/8)$ with the horizontal. The speed of rain is

The velocity vector diagram for both the cases is shown above. In the flrst case, when the man is moving down the incline $\overrightarrow{OA}$ represents the velocity of the man $\left({\overrightarrow{v}}_M\right),\overrightarrow{OC}$ represents the velocity of the rain $\left({\overrightarrow{v}}_R\right)\text{ and }\overrightarrow{AC}$

represents the velocity of the rain relative to the man  $\left({\overrightarrow{v}}_{RM}={\overrightarrow{v}}_R-{\overrightarrow{v}}_M\right)$ Similarly, when the man is moving up

$\overrightarrow{OB}={\overrightarrow{v}}_M,\overrightarrow{OC}={\overrightarrow{v}}_R\text{ and }\overrightarrow{BC}={\overrightarrow{v}}_{RM}$ In the above figure, $OA = OB = 5$ and $\tan \theta =7/8$

Now

$BM=AB\cos {37}^{\circ }=10\cos {37}^{\circ }=8$

and $AM=AB\sin {37}^{\circ }=10\sin {37}^{\circ }=6$

Now, using $\tan \theta =\frac{CM}{BM}=\frac{AC+AM}{BM}$ we have

$\frac{7}{8}=\frac{AC+6}{8}$ gives $AC=1$

In $\mathrm{△}OAC,OA=5,AC=1\text{ and }\mathrm{\angle }OAC={90}^{\circ }+{37}^{\circ }$

Hence, the length of OC, which is the rain speed, equals

$v_R=\sqrt{5^2+1^2-2\times 5\times 1\cos \left({90}^{\circ }+{37}^{\circ }\right)}\mathrm{m}/\mathrm{s}=4\sqrt{2}\mathrm{m}/\mathrm{s}$

Alternate method:

Consider that $V_r=V_x\hat{i}+V_y\hat{j}$, 

In first case man is sliding down the incline 

$V_m=-4\hat{i}-3\hat{j}$

$V_{rm}=(V_x+4)\hat{i}+(V_y+3)\hat{j},$Since this is vertical $V_x=-4m/s$

When the man is moving upwards 

$V_m=4\hat{i}+3\hat{j}$

$V_{rm}=(V_x-4)\hat{i}+(V_y-3)\hat{j},$this makes and angle of ${\tan }^{-1}\left(7/8\right)$ with horizontal

So $$\begin{gathered} \frac{V_y-3}{V_x-4}=\frac{7}{8} \\ \Rightarrow V_y=-4 \end{gathered}$$

Speed of the rain is $4\sqrt{2}$