Electric potential and free fall acceleration problem

[Punchlinegirl]

On planet Tehar, the free fall acceleration is the same as that of Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A 4.05 kg ball having a charge of 4.65 $$\mu C$$ is thrown upward at a speed of 22.2 m/s, and it hits the ground after an interval of 1.86 s. What is the potential difference between the starting point and the top point of the trajectory? Answer in units of kV.

First I drew a free body diagram and found that the only forces on the ball were the force of the charge and gravity.
So qE + mg = ma
where a is constant.
Then I figured out the height of the ball by adding the final and initial velocities and dividing by 2 and multiplying by the time/2.
22.0+ 0 /2 *1.86/2
h= 10.323
Then I used conservation of energy
KE_o + PE_0 +EPE_o = KE_f + PE_f + EPE_f
1/2mv^2_o +0 +0 = 1/2 mv^2_f + mgh +qV
So V= 1/q ((1/2)mv^2_0 -mgh)
V= 1/4.65 x 10^-6 ((1/2)4.05(22.2^2) - 4.05 (9.8)( 10.323))
V=1.26x 10^8
V= 1.26 x 10^4 kV
This isn't right.. can someone tell me what's wrong?

 

[andrevdh]

Well what I did is first calculated the gravitational acceleration of the planet from the data. That enables you to calculate the electric field strength from your first equation. That with the height gave me
$$V=Ed=1.26 \times 10^8\ V$$

 

[Punchlinegirl]

I got that, but the answer is supposed to be in kV. So that would make it 1.26 x 10^4, right?

 

[Punchlinegirl]

Is this the right approach or am I doing it completely wrong?

 

[andrevdh]

Sorry I do not understand the system, since sometimes it notifies me via e-mail when someone posted in the thread, but it did not do it in this case. kilo is $10^3$ therefore $10^5$ will be left after taking away 3 from 8.