Electric Potential related question

[kfan321]

On planet Tehar, the free-fall acceleration is same as that on Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A 2 kg ball having a charge of 5 uC is thrown upward at a speed of 20.1 m/s. It hits the ground after an interval of 4.1 sec. What is the potential difference between the starting point and the top of the trajectory?

Well my plan was to use V = E d
So I need to find E and d and I'll have the problem solved.
I used V2^2 = V1^2 + 2ad
v2 is 0, v1 is 20.1, and i used 9.8 for a.
So d came out to 20.6. But I don't think that's correct because I didnt account for the acceleration due to the electric field? So how do I do that?
To solve for E, I wanted to use F/q but not enough info given. Thought about (KQ)/r^2 but not sure if this scenario can be considered point charges...

 

[Gamma]

Due to the electric field, now effective accelaration is changed. Use Newton's second low to find a. (Hint: What are the two forces now acting on the object?)

 

[kfan321]

Well you really haven't said anything I don't already know. I've said force gravity and electric field will deccelerate the ball... and that's where I am stuck

 

[robphy]

Well you really haven't said anything I don't already know. I've said force gravity and electric field will deccelerate the ball... and that's where I am stuck

Did you draw a free-body diagram?
As you have determined, a is an unknown. d is also unknown. So, your single kinematic equation is not sufficient to find d. Note that you have not used the information about the time of flight. Is there another applicable kinematic equation that can be used here?

 

[kfan321]

i've tried d = v1t + .5 at^2
still have 2 variables. So I did a system of equations with the first kinematic equation. Subbing d into a, hoping to find the new a and it came out to -9.8. lol...

...I'm all for not doing other peoples homework for them, but you guys are hiding the info like it's some top FBI confidential document.

 

[robphy]

i've tried d = v1t + .5 at^2
still have 2 variables. So I did a system of equations with the first kinematic equation. Subbing d into a, hoping to find the new a and it came out to -9.8. lol...

If you do things algebraically, you might find a single equation with one unknown. (That is, there may be a simpler starting point.)

...I'm all for not doing other peoples homework for them, but you guys are hiding the info like it's some top FBI confidential document.

In "Homework, Coursework, & Textbook Questions", you are nudged toward (but not given) the answer... but you have to show your work.

 

[kfan321]

using the 2 kinematic equations previously

0 = v1^2 + 2*a*v1*t + a^2 * t^2

this is what I've done on post #5, i knew everything except a in that equation, so I solved for it and it came out to -9.8. Which can't be right b/c of the field, so where did I go wrong?

 

[robphy]

v2 is 0

Why is this true?

 

[kfan321]

b/c at the top of the trajectory, it's 0 m/s, and I use time as 2.05 instead of 4.1...
But now I think about it, time shouldn't be half... Great, back to square one.

You know what, nvm the thread. I think I am better off figuring it out myself.