Correct Answer - B
From the conservation of energy, we know that the potential energy at height `h` is equal to the kinetic energy at ground.
Therefore, at height `h`, the potential energy of ball `A` is
`m_(A)gh`
Its kinetic energy at ground is
`(1)/(2) m_(A)V_(A)^(2)` `:. m_(A) gh = (1)/(2) m_(A)V_(A)^(2)`
`rArr v_(A) = sqrt(2gh)`
Similarly, `v_(B) = sqrt(2gh)`
Therefore, `v_(A) = v_(B)`
[NOTE : In the question, it is not mentioned that the magnitudes of thrown velocity of both balls are same, which is assumed in the solution].