# Electrical kinematics

1. Jun 26, 2011

### Leeoku

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.

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?

2. Jun 26, 2011

### gdbb

I think you might have misinterpreted the question. The electron's path is parabolic. What is its velocity going to be when it returns to its initial height? Hint: non-zero.

3. Jun 26, 2011

### Leeoku

so... if i assume it is a parabola... lets say it is initially at height H.

It starts going to a max with V initial at V_y. It goes to a max and starts coming down. Once it passes height H again.. will it have the same velocity again?

If so... does that mean for my equation v = v_0+at
-V_y = V_y+at

4. Jun 27, 2011

### gdbb

That's correct! Now solve that equation, and what do you get? :)

5. Jun 27, 2011

### Leeoku

oh thats how the 2 comes about. thanks <3