so hmm check the picture below
[tex]\bf \qquad \textit{initial velocity}\\\\
h = -16t^2+v_ot+h_o \qquad \text{in feet}\\
\\
\begin{cases}
v_o=\textit{initial velocity of the object}\to &64\\
h_o=\textit{initial height of the object}\to &12\\
h=\textit{height of the object at "t" seconds} \end{cases}\\\\
-----------------------------\\\\[/tex]
[tex]\bf \textit{vertex of a parabola}\\ \quad \\
\begin{array}{lccclll}
h(t)=&-16t^2&+64t&+12\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
part 1)
it takes [tex]\bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds[/tex]
part 2)
[tex]\bf \textit{now, doubling }v_o\\\\
\begin{cases}
v_o=\textit{initial velocity of the object}\to &128\\
h_o=\textit{initial height of the object}\to &12\\
h=\textit{height of the object at "t" seconds}\end{cases}\\\\
-----------------------------\\\\
\textit{vertex of a parabola}\\ \quad \\
\begin{array}{lccclll}
h(t)=&-16t^2&+128t&+12\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
it will reach the maximum height at [tex]\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet[/tex]
how much higher than before is that? well, what was the y-coordinate for when the vₒ was 64? what did you get for [tex]\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}[/tex] ?
subtract that from this height when vₒ is 128 or doubled, to get their difference, that's how much higher it became
