Rain is falling at an angle of 30^{circ} from | vedclass

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(B) Let the velocity of rain be $\vec{v}_r$. The horizontal component of rain velocity is $v_{rx} = v_r \sin(30^{\circ})$ and the vertical component i

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) Let the velocity of rain be $\vec{v}_r$. The horizontal component of rain velocity is $v_{rx} = v_r \sin(30^{\circ})$ and the vertical component is $v_{ry} = v_r \cos(30^{\circ})$.Given that the rain is falling at $30^{\circ}$ from the vertical with a speed of $40 \ m/s$, we have $v_r = 40 \ m/s$.Thus, $v_{rx} = 40 \sin(30^{\circ}) = 40 	imes 0.5 = 20 \ m/s$ and $v_{ry} = 40 \cos(30^{\circ}) = 40 	imes \frac{\sqrt{3}}{2} = 20\sqrt{3} \ m/s$.The car is moving in the opposite direction to the horizontal component of the rain with a speed $v_c = 40 \ m/s$.Let the velocity of the car be $\vec{v}_c = 40 \hat{i} \ m/s$. The velocity of rain relative to the car is $\vec{v}_{rc} = \vec{v}_r - \vec{v}_c$.Assuming the rain's horizontal component is in the negative $x$-direction, $\vec{v}_r = -20 \hat{i} - 20\sqrt{3} \hat{j}$.Then $\vec{v}_{rc} = (-20 \hat{i} - 20\sqrt{3} \hat{j}) - (40 \hat{i}) = -60 \hat{i} - 20\sqrt{3} \hat{j}$.The angle $\alpha$ with the vertical is given by $	an(\alpha) = \frac{|v_{rc,x}|}{|v_{rc,y}|} = \frac{60}{20\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.Therefore, $\alpha = 	an^{-1}(\sqrt{3}) = 60^{\circ}$.

Rain is falling at an angle of $30^{\circ}$ from the vertical due to the wind with a speed of $40 \ m/s$. $A$ car is travelling horizontally in the direction opposite to the wind, at a speed of $40 \ m/s$. At what angle from the vertical will it experience the rain falling from (in $^{\circ}$)?

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