We are given that a projectile starts off with initial vertical velocity $V_0$, and reaches maximum height $h$ at constant time $T$.
The height $y$ of the projectile at time $t$ is given by:
$y=V_0t-\frac{1}{2}gt^2$
Then clearly from the first statement:
$h=V_0T-\frac{1}{2}gT^2$
Given that, I'm looking for a function $f(h,t)$ which equals the height of the projectile at time $t$ whose maximum height is $h$. My first thoughts were that:
$\frac{\partial y}{\partial t}=V_0-gt$,
Because the projectile stops at the top:
$V_0-gT=0$
Solving for g:
$V_0=gT$
$g=\frac{V_0}{T}$
I think I can plug g back into $y$ because nothing in it is a function of lowercase $t$?:
$y=V_0t-(\frac{V_0}{T})t^2$
I don't know if that works, or not. If it does, I can solve for $T$ in $h=V_0T-\frac{1}{2}gT^2$ and plug it in, and then I'm done.
EDIT
That final solve would ensure $g$ is still in the equation.
The altitude is given by $$y (t)=-\frac {1}{2}gt^2+V_0t $$
and the velocity by $$V (t)=\frac {dy}{dt}=-gt+V_0$$
the maximum height $h $ is reached when $V (T)=0$
thus $$T=\frac {V_0}{g} $$ and $$h=y (T)=\frac {1}{2}\frac {V_0^2}{g}$$