Ball rolling down a segment then up a parabolic segment

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talaroue
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Homework Statement


A 181 g ball is released from rest H=5.19 m above the bottom of the track shown in Figure. It rolls down a straight 21° segment, then back up a parabolic segment whose shape is given by y=x2/4, where x and y are in m. How high will the ball go on the right before reversing direction and rolling back down?
Diagram of a ball released from rest on a straight incline at angle θ, height H above a dashed horizontal reference line, which then curves up into a parabolic track on the right.


Homework Equations


Ki+Ui=Kf+Uf
K=mv^2/2


The Attempt at a Solution


I am not sure how to start this because everytime i go to one direction i hit a dead end.
At the beginning it has both Ki=0 and Ui=0 correct?
Then at the final it has Kf and Uf right?
 
Here is my work but I don't even think I am on the right track...

This is for rolling down incline
Handwritten physics work on paper, rotated sideways: sketch 1 shows a ball on an incline at 21 degrees, equation mgy = mv^2/2, and a derivation v = sqrt(2 gy) giving v = 6.04 m/s; sketch 2 shows a ball on a curved ramp at 2.2 degrees.



This is for rolling up the parabola
Handwritten physics work showing a ball rolling up a curved parabolic track with velocity v. Equations: KE = mv^2/2, Up = mgy = mg x^2/4, mv^2/2 = mg x^2/4, x = sqrt(2mv^2/mg) = 2.73m, and y = 2.73^2/2 = 3.72m. Top line reads v = 6.04 m/s.

EDIT: I put the wrong dominator in for the very last part when calculating Y it should be 4, but that anwser did not work either
 
WOOOOOOW I am so sorry for wasting everyones times the parabola made me over think this equation since Ki+Ui=Kf+Kf...energy would force the ball up the same height as it began.
 
The question gave you additional information to deceive you. Realizing this in the end only makes you understand energy conservation better. It's hardly wasting anyone's time.
 
Yeah, I see what you mean but I didn't need to make a thread after.