To a man going with a speed of `5 ms^-1`, rain appears to be falling vertically. If the actual speed of rain is `10 m s^-1`, then what is the angle made by rain with the vertical ?

Answer

`30^@`

Explanation

`vec v_m = 5 hat i m//s, vec v_(r//m) = -v hat j, vec v_r = vec v_(r//m) + vec v_m = 5 hat i - v hat j`
`v_r = sqrt(5^2 + v^2)) = 10 m//s`
`rArr v = 5 sqrt(3)) m//s`
So `vec v_r = 5 hat i - 5 sqrt(3)) hat j m//s`
`tan theta = (5)/(5 sqrt(3)) rArr theta = 30^@`.
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