To a man walking at the rate of `3 km//h` the rain appear to fall vetically douwnwards. When he increases his speed `6 km//h` it appears to meet him at an angle of `45^@` with vertically. Find the speed of rain.

Explanation

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

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