Electrical kinematics

[Leeoku]

Homework Statement

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

Homework Equations

v = v_0+at
F = qE = MA

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 because 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?

 

[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.

 

[Leeoku]

so... if i assume it is a parabola... let's 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

 

[gdbb]

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

 

[Leeoku]

oh that's how the 2 comes about. thanks <3