A particle `A` is projected with speed `V_(A)` from a point making an angle `60^(@)` with the horizontal. At the same instant, second particle `B` (lie in the same horizontal plane) is thrown vertically upwards from a point directly below the maximum height point of parabolic path of `A`, with velocity `V_(B)`. if the two particles collide then the ratio of `V_(A)//V_(B)` should be:

Options

A. 1
B. `2/sqrt(3)`
C. `sqrt(3)/2`
D. `sqrt(3)`

Answer

B

Explanation

At maximum height vertical component of velocity becomes zero.
`v^(2)=u^(2)+2as`

For A : `0=v_(A)^(2)sin^(2)60^(@)-2gh`
`2gh=v_(A)^(2)sin^(2)60^(@)=v_(A)^(2)(3//4)`
`v_(A)=sqrt((8gh)/3)`
For B : `0=v_(B)^(2)-2gh`
`v_(B)=sqrt(2gh), v_(A)/v_(B)=2/sqrt(3)`