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

Explanation

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

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