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To a man walking at the rate of 3 km/hr the rain appears to fall vertically. When he increases the speed to 6 km/hr it appears to meet him at an angle of $45^{\circ}$ with vertical. Find the speed of rain.

$(a)\;3 km/hr\quad (b)\;\frac{3}{\sqrt 2} km/hr \quad (c)\;5km/hr \quad (d)\;3 \sqrt 2 km/hr$

Let $\hat i$ and $\hat j$ be unit vector in horizontal and vertical respectively.
Let velocity of rain $v_r=a \hat i+b \hat j$
speed of rain$=\sqrt {a^2+b^2}$
Case I:
when relative velocity of rain w.r.t man is vertical
$v_{rm}=\bar v_r-\bar v_m$
$v_m=3 \hat i$
$v_{rm}=(a-3) \hat i+b \hat j$
Since $v_{rm}$ is vertical
$a-3=0=>a=3$
Case II:
when relative velocity is at $45 ^{\circ}$
$\bar v_m=6 \hat i$
$v_{rm}=(a-6) \hat i+b \hat j$
$\qquad=- 3 \hat i+b \hat j$
Since $\tan \theta=\large\frac{b}{-3}$
tan 45=1
=> $|b|=3$
Therefore $Speed=\sqrt {3^2+3^2}=3 \sqrt 2$
Hence d is the correct answer.
edited May 24, 2014