Correct Answer - C
Let ` vec u` be the velocity of man anf ` vec v` be the velocity of tain. If ` hat i` and ` hat j` are unit vectors along horizontal road and vertical direction respectively. According to question
` vec u = 8 hat i` abd ` ` vec v= v_x hat i+ vec v_y hat j`
` If ` vec v_(rm)` is the velocity of rain relative to man, then ltbr. ` vec v_(rm) = vec v- vec u = (v_x hat i = v_y hat j) - 8 hat j`
` = (v_x - 8 ) hat i+ v_y hat j`
As rain appears to be falling vertcally downwards.
` vec v(rm) =- v-r hat j` :. ` - vec v-r hat j = (v_x - 8) hat + v_y hat j`
Comparing the coefficients of ` hat i` and ` hat j` on both the sides, we get
` v_x - 8 =0` or `v_x =8 ` and ` -v-r =v_y`
When man incrases his speed, then ` vec u = 12 hat i`.
Let the relative velocity of rain w.r.t. man be ` vec v(r` m`) the
` vec v(r`m) = vec v- vec u= (v_x hat i+v_y hat j) - 12 hat j`
` = 8 hat i + v_y) - 12 hat i =- 4 hat i + v_y hat j`
As rain appears to be falling at an angle ` 30^@` with the vertical, therefore angle with road i.e. horezontal derection ` theta= (90^@ - 30^@) = 60^@`. Now `
` tan 60^@ = v_y/(-4) ` or ` v_y =- 4 tan 60^@ =- 4 sqrt 3 m//s`
` :. vec v= v_x hat i +v_x hat i +v_y hat j= 8 aht -4 sqrt 3 hat j`
` | vec v| = sqrt ((8)^2 + (-4 sqrt 3)^2) = sqrt(64 + 48)`
`= 10. 58 km h^(-1)`.
