To a man running at the rate of 4km/h, the rain appears to fall vertically. When he increases his speed to 8km/h, rain appears coming to man at an angle of 45∘ with vertical. Find the magnitude of velocity of rain with respect to ground.
A
3√2km/h
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B
4√2km/h
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C
2√2km/h
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D
5√2km/h
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Solution
The correct option is B4√2km/h Let the velocity of the rain w.r.t ground be →vrg=a^i+b^jkm/h
where ^i and ^j are the unit vector in horizontal and vertical direction respectively.
Magnitude of velocity of rain w.r.t ground |→vrg|=√a2+b2km/h...(1)
In first case,
Velocity of man w.r.t ground →vmg=4^ikm/h →vrm=→vrg−→vmg=(a−4)^i+b^j ∵Rain appear to man to be in vertical direction, x−component of →vmg should be zero. ⇒(a−4)=0⇒a=4
In second case; →vmg=8^ikm/h →vrm=(a−8)^i+b^j (but a = 4) →vrm=−4^i+b^j
In this case, rain appears to be at 45o with vertical. tan45∘=b−4⇒b=−4 |b|=4
Magnitude of velocity of rain w.r.t ground |→vmg|=√a2+b2 ∴|→vmg|=√42+42=4√2km/h