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)`

Options

A. `10m//s`
B. `20m//s`
C. `15m//s`
D. `30m//s`

Answer

A

Explanation

Rain is falling …………….
w.r.t. observer
let final velocity of rain be v'
initial of rain `= (3v)/(5)hati - (4v)/(5)hatj`

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