    Question

# A man running on a horizontal road at 8 km/h finds the rain falling vertically. He increases his speed to 12 km/h and finds that the drops make angle 30∘ with the vertical. Find the speed and direction of the rain with respect to the road.

A

vrain=47 km/hr

α=tan1(32)

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B

vrain=5 km/hr

α=cot1(32)

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C

vrain=47 km/hr

α=tan1(32)

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D

vrain=4133 km/hr

α=tan1(23)

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Solution

## The correct option is C vrain=4√7 km/hr α=tan−1(√32) Given vman=8^i Rain appears to him to be falling vertically ⇒vr/m=−b^j vrm=vr−vm⇒vr=vrm+vm ⟶ Velocity of rain vector is the resultant of vrm & vman vector Let's draw vector diagram vm=8 Let velocity of rain be a and let it fall at angle α w.r.t vertical ⇒in unit vector notation vr sin α^i−vrcos α^j a sin α^i−acos α^j ⇒vrm=vr−vm ⇒−b^j=(a sin α−8)^i−a cos α^j⇒ a sin α−8=0⇒sin α=+8a ...............(i)eqn Now question says vman=12^i vr=a sin α^i−acos α^j As rain's velocity vector won't change Substituting equation (iv) in (iii) we get a cos α=8√32 .............(v) From equation (i) asin α=8 Squaring and adding a2=64+64×34 a2=64(74) a=4√4 ⇒(vr=4√7) km/hr Dividing (i) & (v) tan α=√2√3 α=tan−1(2√3) Rain appears to be falling at 30∘ with vertical That means he's talking about vrm Hmmmm...... vrm=vr−vm Let's draw vector diagram and see Now the magnitude would have changed. Let's assume |→vrm| to be c vrm=−c2^i−c√32^j vrm=vr−vm −c2^i−c√32^j=(asin α−12)^i−a cos α^j ⇒(a sin α−12)=−c2 .........(ii) & a cos α=c√32 ...............(iii) From equation (i) we know sin α=8a substituting in equation (ii) we get a×8a−12=−c2 −4=−c2⇒c=8 ................(iv)  Suggest Corrections  0      Similar questions  Related Videos   Relative Motion in 2D
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