To a man walking at the rate of 3 km/h the rain appears to fall vertically. When be increases his spee to 6 km/h it appears to meet him at an angle of `45^(@)` with vertical. Find the speed of rain.

Explanation

Let `hati` and `hatj` be the unit vectors in horizontal and vertical directions respectively.

Let velocity of rain `v_(r )=ahati+bhatj`
Then speed of rain will be `|v_(r )|= sqrt(a^(2)+b^(2))` In the first case `v_(m)` =velocity of man=`3hati`
`therefore v_(m)=v_(r )-v=(a-3)hati+bhatj`
It seems to be in vertical direction. Hence, a-3=0 or a=3
In the second case `v_(m)=(a−6)hati+bhatj`
This seems to be at `45^(@)` with vertical. Hence, |b|=3
Therefore, from Eq. (fi) speed of main is `|v_(r )|=(sqrt(3)^(2)+(3)^(2))=3sqrt(2)km//h`