A girl riding a bicycle with a speed of ` 5 m//s` to wards Noth direction, observes rain falling vertically down. If she increases her speed to `10 m//s`, rain appeard to meet her at ` 45^@` to the vertical . What is the speed ot the rain ? In what direction does rain fall as observed by a ground based observer ?

Explanation

Assume north to be `hat(i)` direction and vertically downward to be -`hat(j)`.
Let the rain velocity `v_(r)` be a `hat(i) +b hat(j)`
case I Given velocity of girl `=v_(g)=(5m//s)hat(i)`
Let `v_(rg)`=Velocity of rain w.r.t. girl
=`v_(r)-v_(g)=(ahat(i)+bhat(j))-5hat(i)`
`=(a-5)hat(i)+bhat(j)`
According to question rain, appears to fall vertically downward.
Hence, `" " a-5=0rArra=5`
case II Given velocity of the girl , `v_(g)=(10m//s)hat(i)`
`:. " " v_(rg)=v_(r)-v_(g)`
`" " =(a hat(i) + bhat(j))-10hat(i) = (a-10)hat(i)+b hat (j)`
According to question rain appears rain appears to fall at `45^(@)` to the vertical hence tan `45^(@) = (b)/(a-10)=1`
`rArr " " b = a - 10 = 5 - 10 = -5`
Hence, velocity of rain `=ahat(i) + b hat(j)`
`rArr " " v_(r)=5hat(i) - 5hat(j)`
Speed of rain `=|v_(r)|=sqrt((5)^(2)+(-5)^(2))=sqrt(50)=5sqrt(2) m//s`