A man running on a horizontal road at `6 km//h` finds the rain falling vertically. He doubles his speed and find that the raindrops make an angle `37^(@)` with the vertical. Find the velocity of rain with respect to the ground.

Explanation

Let the velocity of rain w.r.t. the ground is `u`.

Recall: `v_(r//m)` is resultant of `v_(r//g)` and `v_(m//g)` (in opposite direction).

`sin theta-6=0`
`u sin theta=6......(i)`
Now the man doubles his speed and `v_(r//m)` makes `37^(@)` with vertical, it may be

Recall: Resultant is towards larger vector.


`tan 53^(@)=(usin(90+ theta))/(12 + ucos(90+ theta))`
`4/3=(ucos theta)/(12-usin theta)=(u cos theta)/(12-6)`
`ucos theta=8..........(ii)`
`u=sqrt((usin theta)^(2)+(ucos theta)^(2))`
`=8sqrt((6)^(2)+(8)^(2))=10km//h`
`tan theta=(usin theta)/(ucos theta)=6/8=3/4 implies theta=37^(@)`
`v_(r//g)=10 km//h` at an angle `37^(@)` with vertical direction