Check the picture below.
so then, the highest point will be at the y-coordinate of its vertex.
[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+48}t\stackrel{\stackrel{c}{\downarrow }}{+4} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]
[tex]\bf \left( \qquad ,~~4-\cfrac{48^2}{4(-16)} \right)\implies \left( \qquad ,~~4+\cfrac{2304}{64} \right)\implies (\qquad ,~4+36) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\qquad ,~~\stackrel{\stackrel{\textit{how high}}{\textit{it went}}}{40})~\hfill[/tex]