Question

When a man moves down an inclined plane with a constant speed of $5m{s}^{-1}$ , 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 $\theta ={\tan }^{-1}\left(\frac{7}{8}\right)$ with the horizontal. If the angle of the inclined plane with the horizontal is $37^{\circ }$ , 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)

${\vec{V}}_{Rm}={\vec{V}}_R-{\vec{V}}_m$
${\vec{V}}_{Rm}=\left(V_x\hat{ı}-V_y\hat{ȷ}\right)-(4\hat{ı}-3\hat{ȷ})$


Rain is falling vertically. Therefore,
when the man is going up
${\vec{V}}_{Rm}={\vec{V}}_R-{\vec{V}}_m=\left(4\hat{ı}-V_y\hat{ȷ}\right)-(-4\hat{ı}+3\hat{ȷ})<br/>{\vec{V}}_{Rm}=8\hat{i}-\left(V_y+3\right)\hat{j}<br/><br/>$

Given,
$\tan \theta =\frac{V_y}{V_x}=\frac{Vy+3}{8}=\frac{7}{8}$
$V_y=4,V_R=\sqrt{v_x^2+v_y^2}=\sqrt{32}\mathrm{m}{\mathrm{s}}^{-1}$