An elevator at rest which is at 10th floor of a building is having a plane mirror fixed to its floor. A particle is projected with a speed `(sqrt(2))m//s` and at `45^(@)` with the horizontal as shown in the figure. At the very instant of projection, the cable of the elevator breaks and the elevator starts falling freely. what will be the separation between the particles and is image 0.5 s after the instant of projection?

Options

A. `0.5 m`
B. `1 m`
C. `2 m`
D. `1.5 m`

Answer

B

Explanation

`u_(y)=sqrt(2) sin 45^(@)=1m//s`
In vertical direction,
`s_(1)=`displacement of partical
`=(1)(0.5)-(1/2)(g t)^(2)=0.5-(1/2) (g t)^(2)`
`s_(2)` =displacement of mirror
`=-(1/2)g t^(2)`
`:.` Vertical distance of partical from nirror,
`s=s_(1)-s_(2)=0.5 m`
Hence, distance between particle and its image
`=25=1 m`.