In this question $g=9.8$ (acceleration of free fall). You are also given that when $x=0$ $v=0$.
My answer is $v^2=400g(1-e^\frac{x}{200})$. I obtained it by integrating both sides so that $x=-200\ln(g-0.0025v^2) + A$ [where $A$ is the constant of integration]. Substituting boundary conditions to obtain $A$ yields $A =200lng$.
This is how the other answer was obtained (uploaded it from the mark-scheme of the exam paper):
Its clear what the problem here is: it depends which side of the equation you choose to place the constant of integration. How can this be?
By the way here is the exam question:
I'd really appreciate it if someone could explain what it is i'm doing wrong.
Thank you.