Rain is falling with constant and uniform velocity `v` at angle `37^(@)` vertical. An observer starts from rest and moves with uniform acceleration `2 m//s^(2)` in horizontal direction (with the rain). After `6 sec`. Velocity of rain drop again appear at angle `37^(@)` from vertical. If velocity of rain fall `w.r.t.` ground is `5n`, then `n` is : `(tan37^(@) = 3//4)`

Answer

2

Explanation

`w.r.t` observer
let final velocity of rain be `v'`
initial velocity of rain `= (3v)/(5)hat(i) - (4v)/(5)hat(j)`

acceleration of rain `= -2 hat(j) m//s^(2)`
final velocity of rain `= -(3v')/(5)hat(i)- (4v')/(5)hat(j)`
in `y-`direction
`v_(y "final") = v_(y "initial")`
`v' = v`
in `x-`direction
`v_(y "final") = v_(y "initial") + at`
`-(3v')/(5) = (3v)/(5) -2^(***)6`
`-(3v)/(5) = (3v)/(5) - 12`
`v = 10 m//s`