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

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#### Solution

V_{rg} is the velocity of rain that appears to the girl.

We must draw all vectors in the reference frame of the ground-based observer.

Assume north to be `hati` direction and vertically downward to be `(-hatj)`.

Let the rain velocity `v_r = ahati + bhatj`'

**Case I:** According to the problem, velocity of girl `v_g = ((5m)/s)hati`

Let `v_(rg)` = velocity of rain w.r.t. girl

= `v_r - v_g`

= `(ahati + bhatj) - 5hati`

= `(a - 5)hati + bhatj`

According to the question, rain appears to fall vertically downward.

Hence, `a - 5 = 0 ⇒ 4a = 5`

**Case II:** Now velocity of the girl after increasing her speed,

`v_g = ((10m)/s)hati`

∴ `v_(rg) = v_r - v_g`

= `(ahati + bhatj) - 10hati`

= `(a - 10)hati + bhatj`

According to the question, rain appears to fall at 45 to the vertical.

Hence `tan 45^circ = b/(a - 10)` = 1

⇒ `b = a - 10`

= `5 - 10`

= `- 5`

Hence, velocity of rain = `ahati + bhatj`

⇒ `v_r = 5hati - 5hatj`

Speed of rain

= `|v_r| = sqrt((5)^2 + (-5)^2`

= `sqrt(50)`

= `5sqrt(2)` m/s.

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