Question is simple. A ball is thrown upwards with initial velocity $v_0$. How high does it go up? Assume down is negative and up is positive and no air resistance.
KE at bottom = PE at the top: $\frac12 m \times v_0^2 = m (-9.8)\times h$ which in turn implies $h$ will be negative.
So question is we assumed up is positive and why are we getting a negative height.
My guess is that we are dealing with energy which is a scalar quantity and hence signs should not be used. Not sure of this though.
The error is with your sign in the potential energy, or alternatively with your sign for $g$. Your right hand side $mgh$ should increase as $h$ increases, but with $g=-9.8$ the potential energy will decrease as you increase the height. In other words, your expression for PE should be $-mgh$ if you use $g=-9.8$ and you take vertical up to be increasing heights.
The potential energy difference between point $A$ and $B$ is defined in terms of the work of a conservative force ${\bf F}$ on any path from $A$ to $B$: $$ U(A)-U(B) = W_{AB} = \int_A^B {\bf F} \cdot {\mathrm d}{\bf r}. $$ If ${\bf F} $ is a uniform force (independent of the position) and the $z$ axis is aligned with the force $$ W_{AB} = {\bf F} \cdot ({\bf r}(B) - {\bf r}(A)) = F_z (z(B)-z(A)) $$ Therefore, if the force points downwards ($F_z = -mg$), the work to go from a point at height $z(A)$ to another point at a higher height $z(B)=z(A)+h$ ($h>0$) is $$ W_{AB} = -mgh$$, a negative quantity because the displacement is opposite to the force. Thus, if we chose $U(A)$ as our reference potential energy, $$ U(B) = -W_{AB} = +mgh $$ When we apply the conservation of energy principle between point $A$, where the potential energy has been taken zero, and point $B$, where the kinetic energy vanishes, we have $$ E =K(A)= \frac12 m v^2(A) = U(B) = mgh. $$ Notice that the same result, with the same sign, would be obtained if the direction of the $z$ axis were chosen in the reversed direction. Indeed, the component of the force would change its sign, but the height difference will do the same. In a way, the scalar nature of the work ensures its invariance with respect to a reflection in a $x-y$ plane. However, this is not equivalent to saying that signs should not be used. If the height of point $B$ is lower than point $A$ ($z(B)=z(A)-h$, with $h>0$), $U(B)=-mgh$, again independently of the choice of the axis direction.
Lets look at the equation of motion
case I
$$m\ddot h=-m\,g$$ the solution with the initial conditions $~h(0)=0~,\dot h(0)=v_0~$ is
$$h(t)=v_0\,t-\frac g2\,t^2$$
from here to obtain $~h_{\rm max}~$ you solve this two equations
$$h(t)=h_{\rm max}~,\dot h=0$$
for $~t_{\rm max}~$ and $~h_{\rm max}~$
$$h_{\rm max}=\frac 12 \frac {v_0^2}{g}\quad, t_{\rm max}=\frac {v_0}{g}\quad, g=9.81$$
case II
$$m\ddot h=-m\,(-g)=m\,g$$
the solution
$$h(t)=v_0\,t+\frac g2\,t^2$$
thus
$$h_{\rm max}=-\frac 12 \frac {v_0^2}{g}\quad, t_{\rm max}=-\frac {v_0}{g}\quad, g=9.81$$
but the time $~t_{\rm max}~$ can not be negative ! this mean that g negative is wrong assumption
