Two small balls having the same mass and charge and located on the same vertical at heights `h_(1)` and `h_(2)` are thrown in the same direction along the horizontal at the same velocity v. The first ball touches the ground at a horizontal distance R from the initial vertical position. At what height `h_(2)` will the second ball be at this instant? Neglect any frictional resistance of air and the effect of any induced charge on the ground.

Answer

`H_(2)=h_(1)+h_(2)-g(l/v)^(2)`

Explanation

Time to fall first ball
`t=l/v :. h_(1)=1/2xxgxxl^(2)/v^(2)`
`:.` Height of 2nd ball
`=h_(2)-1/2gl^(2)/v^(2)`
`pr=h_(2)+1/2 g l^(2)/v^(2), h_(2)+h_(1)-g (l^(2)/v^(2))`