To a person going westwards with a speed of `6 km h^(-1)`, rain appears to fall vertically downwards with a speed of `8 km h^(-1)`. Find the actual speed of the rain and its direction.

Explanation

Refer to 2 (c ). 54, velocity of person,
` vec v_P = vec (OA)`, where
` u_P=6 km h^(-1)` Relatice velocity of rain w.r.t. to person,
` vec v_(rP) = vec v_r - vec v_P or vec _r = vec _(rP) + vec v_P`
.
or `vec v_r = vec (OB) + vec (OA) = vec (OC)`
:. Magnitude of `vec v_r`
` |vec (OC)| = sqrt ((v_(rP))^2+ (v_(P))^2)`
` = sqrt(8^2 =- 6 ^2 ) = 10 km h^(-1)`
Let ` theta` be the angle which `vec v_r` makes with the vertical, then
` tan theta = (BC)/(OB) = (OA)/(OB) = u_p/v_(rP) = 6/8 = 0.75`
` theta = 36^@ 52' (east of vertical).