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Question

(B) Let the velocity of the rain with respect to the ground be $\vec{V}_r = V_x \hat{i} - V_y \hat{j}$ and the velocity of the man be $\vec{V}_m = V_m

Vedclass · Accepted Answer

$\frac{20}{\sqrt{3}}\,m/s$

Vedclass · Answer

(B) Let the velocity of the rain with respect to the ground be $\vec{V}_r = V_x \hat{i} - V_y \hat{j}$ and the velocity of the man be $\vec{V}_m = V_m \hat{i}$.When the man is running,the relative velocity of rain with respect to the man is $\vec{V}_{rm} = \vec{V}_r - \vec{V}_m = (V_x - V_m) \hat{i} - V_y \hat{j}$.Since the rain appears to fall vertically to the running man,the horizontal component of $\vec{V}_{rm}$ must be zero: $V_x - V_m = 0$,so $V_x = V_m$.The vertical component of the relative velocity is given as $V_y = 10\,m/s$.When the man stops,the rain appears to fall at an angle of $30^{\circ}$ with the vertical. Thus,$	an 30^{\circ} = \frac{|V_x|}{|V_y|}$.$\frac{1}{\sqrt{3}} = \frac{V_x}{10} \implies V_x = \frac{10}{\sqrt{3}}\,m/s$.The speed of the rain with respect to the ground is $V_r = \sqrt{V_x^2 + V_y^2} = \sqrt{(\frac{10}{\sqrt{3}})^2 + 10^2} = \sqrt{\frac{100}{3} + 100} = \sqrt{\frac{400}{3}} = \frac{20}{\sqrt{3}}\,m/s$.

$A$ person running horizontally observes that rain is falling on his head vertically with a speed of $10\,m/s$. When he stops,he observes that the rain is falling at an angle of $30^{\circ}$ with the vertical. Find the speed of the rain with respect to the ground.

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