Question

When a man moves down an inclined plane with a constant speed of $5ms^{- 1}$ , 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. $5 \, m \, s^{- 1}$
• D. $\sqrt{73} \, m \, s^{- 1}$

Answer 1

Answer: (B)

$\vec{V}_{R m}=\vec{V}_R-\vec{V}_m $
$\vec{V}_{R m}=\left(V_x \hat{\imath}-V_y \hat{\jmath}\right)-(4 \hat{\imath}-3 \hat{\jmath})$


Rain is falling vertically. Therefore,
when the man is going up
$\vec{V}_{R m}=\vec{V}_R-\vec{V}_m=\left(4 \hat{\imath}-V_y \hat{\jmath}\right)-(-4 \hat{\imath}+3 \hat{\jmath})
\vec{V}_{R m}=8 \hat{i}-\left(V_y+3\right) \hat{j}

$

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