A girl riding a bicycle with a speed of5 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 at45^@ to the vertical . What is the speed ot the rain ? In what direction does rain fall as observed by a ground based observer ?

A girl riding a bicycle with a speed of5 m//s to wards Noth direction,

1.	A girl riding a bicycle with a speed of5 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 at45^@ to the vertical . What is the speed ot the rain ? In what direction does rain fall as observed by a ground based observer ?
Answer» `5sqrt(2)ms^(-1)` `5 m s^(-1)` `10sqrt(2) m s^(-1)` `10 m s^(-1)` SOLUTION :Assume north to be `hati` direction and vertically upwards to be `HATJ` direction. let the velocity of rain be `vecv_(r) = a hati + B hatj` In the first case, `vecv_(g) `= velocity of girl ` = 5hati` `thereforevecv_(rg) = vecv_(r) - vecv_(g) = (a hati + b hatj) - 5hati = (a-5)hati + bhatj` Since rain appears to fall vertically downwards, `THEREFORE a - 5 = 0 or a = 5` In the second case, `vecv_(g) = 10 hati` `thereforevecv_(rg) = vecv_(r)-10 hati = (ahati + bhatj) - 10hati = (a -10) hati + bhatj = - 5hati + b hatj` Since rain appears to fall at `45^(@)` with the vertical, Since rain appears to fall at `45^(@)` with vertical, `therefore a = b = -5` `therefore` The velocity of the rain is `vecv_(r) = 5hati - 5hatj` Speed of the rain is `\|vecv_(r)\| = sqrt((5)^(2) + (-5)^(2)) = 5sqrt(2) ms^(-1)`