A particle is unchanged and is thrown vertically upward from ground level with a speed of `5sqrt(5)` m//s in a region of space having uniform electric field As a result, it attains a maximum height h. The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of `13 m//s` Finally the particle is given a negative charge -q Ignoring air resistance determine the speed (in m//s) with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the same maximum height h.

Answer

`0009`

Explanation

In uniform electric in vertical direction if (+ve) charge feels extra acceleration in downward direction then (-ve) charge will feel acceleration in upward direction.
`v_(uncharged)=5sqrt(5)m//sec`
`v=0, h=height`
`v^(2)-u^(2)=-2(g) h`
`-(5sqrt(5))=-2gh`
`u_(q+)=13m//sec`
`v=0 h = h`
`v^(2)-u^(2)=2(g+F_(E)/m)h`
`0-(13)^(2)=-2(g+(F_(E))/m)h`
Let `u_(q-)=u(say)`
`v=0 h=ht`
`v^(2)-u^(2)=-2(g-(F^(E))/m)h`
`-u^(2)=-2(g-(F_(E))/m)h,u=9m//sec`