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 `hat(i)` and `hat(j)` be the unit vectors in horizontal and vertical directions respectively.

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