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A man running on the horizontal road at ` 8 km h^(-1)` find the rain appears to be falling vertically. He incresases his speed to `12 km h^(-1)` and find that the drops make angle ` 30^2` with the vertical. Fin dthe speed and direction of the rain with respedt to the road.

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We have `vecv_(rain,road)=vecv_(rain,man)+vecv_(man,road)`………i ltbr. The two situations givenin the problem may be represented by the following ure. a. `vecv_(rain,road)` is same in magnitude and direction in bnogh the ures.
Taking horizontal components in equation i. for ure.
`v_(rain,road) sin alpha= 8km/h` ............ii.
Now consider ure. Draw a line `OA_|_ v_(rain, man)` as shown.
Taking components in equation i. along the line OA. `v_(rain, road) sin (30^0+alpha)=12 km/h cos 30^0`..........iii
From ii. and iii
`(sin(30^0+alpha))/(sin alpha)=(12xxsqrt13)/sqrt8xx2)`
`or, `(sin 3^0cosalpha+cos30^0 siN/Alpha)/(siN/Alpha)=(3sqrt(3)/4`
or `1/2 cot alpha + sqrt(3)/2=(3sqrt3)/4`
or, `cot alpha = sqrt3/2
or, ` alpha = cot^-1 sqrt3/2`
From ii, `v_(rain,road)=(8km/h)/(siN/Alpha)=4sqrt7 km/h`.
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