A girl riding a bicycle with a speed of ` 5 m//s` to wards North direction, observes rain falling vertically down. If she increases her speed to `10 m//s`, rain appeared to meet her at ` 45^@` to the vertical . What is the speed of the rain ?

Explanation

Let us draw the vector diagram of the information given and find a and b. we may draw all vectors in the reference frame of ground based observer. . Assume north to be `hati` direction and vertically downwald to be `-hatj`
Let the rain velocity `v_(r)` be `ahati+bhatj`
`v_(r)=ahati+bhatj`
case I given velocity of girl
`=v_g(5(m)/(s))hati`
Let `v_(rg)=` velocity of rain w.r.t girl
`v_r-v_g=(ahati+bhatj)-5hati`
`=(a-5)hatj+bhatj`
Accoding to question rain appears to fall vertically downward Hence `a-5=0impliesa=5`
`v_g=(10(m)/(s))hati`
`v_(rg)=v_r-v_g`
`=(ahati+hatj)-10hati=(a-10)hati+bhatj`
According to question rain appears to fall at `45^@` to the vertical Hence `tan46^@=(b)/(a-10)=1`
`impliesb=a-10=5-10=-5`
Hence, velocity of rain`=ahati+bhatj`
`impliesv_r=5hati-5hatj`
Speed of rain `=|v_r|=sqrt((5)^(2)+(-5)^(2))=sqrt(50)=5sqrt(2)(m)/(s)`