1. The problem statement, all variables and given/known data An electron is projected at an angle of 30.8° above the horizontal at a speed of 8.20×105 m/s in a region where the electric field is E = 388j N/C. Neglecting the effects of gravity, calculate the time it takes the electron to return to its initial height. Answer: 1.23e-08 s 2. Relevant equations v = v_0+at F = qE = MA 3. The attempt at a solution Components of Velocity V_x = Vcos30.8 V_y = Vsin30.8 F = qE = ma a = qE/m = 1.6e-19*388/9.11e-31 = 6.81e13 V = v_0+at (set V = 0 cuz max height) 0 = v_y+at t = 6.17e-9 Now i was playing with nums and found that if i multiplied V_y by 2 then divided it by a that is the right answer, why?