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

Refer Fig. 2 (EP). 30`,
Let velociy of rain be , ` vec v_r = a hat I + b hat j` ….(i)
Ist case : Velocity of girl, ` v_g= ( 5 ms^(-1) hat i`
Velocity of rain w.r.t. girl, ` vec v_(rg) = vec v_g = ( ahat I + b hat j) - 5 hat I = ( a- 5 ) hat + b hat j`
since, rain appeasrs to fall vertically downwards, so `(a - 5) = 0 or ` a= 5 `.
` 2nd csoe : ` vec v_g = (1 0 ms^(-1)) hat i` ltbRgt :. ` vec _(rg) = vec v_r - 10 hat i = (a hat i + b hat j) - 10 aht i = () a - - 10) hat j`
Since, rain appears to fall at ` 45^@` to the vertical, vertical therefore , ` tan 45^@ = b/ ( a- 10 ) = b(5 - 10)/b/( -5)`
or ` b=- 5 tan 45^@ =- 5 1 =- 5`
` vec v_r 5 hat - 5 hat j ` and ` | vec v_r | = sqrt ((5)^2 + (-5)^@)= 5 sqrt 2 ms^(-1)`
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