A particle is uncharged and is thrown vertically upward from ground level with a speed of 25.0 m/s. 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 30.0 m/s. The electric potential at the height h exceeds the electric potential at ground level. Finally, the particle is given a negative charge –q. Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity. I figured out h with Vf^2 = Vi^2 + 2ah (38.55m) I also know difference in energy = mv^2/2 - mv^2/2 = 275 J I also know that at the top, its positively charged, and that the speed required for -q will be less how can i solve the rest?