so hmm check the picture below
[tex]\bf \qquad \textit{initial velocity}\\\\ h = -16t^2+v_ot+h_o \qquad \text{in feet}\\
\\
v_o=\textit{initial velocity of the object}\\
h_o=\textit{initial height of the object}\\
h=\textit{height of the object at "t" seconds}\\\\
-----------------------------\\\\
h(t)=3+80t-16t^2\iff h(t)=-16t^2+80t+3[/tex]
a)
well, clearly is 80 ft/s
b)
when t = 1? well 80(1)
c)
in the picture, x-axis is the time and y-axis is the height
so, it reaches its maximum at the vertex, after "x" seconds
[tex]\bf \begin{array}{lcccll}
h(t)=&-16t^2&+80t&+3\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
so it reached the vertex after [tex]\bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds[/tex]
d)
the maximum height of the ball is [tex]\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet[/tex]
