A particle A is projected with speed `v_(A)` from a point making an angle `60^(@)` with the horizontal. At the same instant, a second particle B is thrown vertically upwards from a point directly below the maximum hieght 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)`