Correct Answer - A::C
Let v is the horizontal velocity of platform in opposite direction. Then from momentum conservation in opposite direction we have,
`(60+40)v=(1)(10cos45^@)`
`:. v=10/100cos 45^@m//s`
`=10/sqrt2cm//s`
`:.` Displacement of platform `=vt`
`=((v)(2usin45^@))/(g)`
`=(10/sqrt2)=(2xx10xx1/sqrt2)/(10)`
`=10cm`