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Answer :
Check the picture below.
so let's simply check about what's its vertex then
[tex]\textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+40}t\stackrel{\stackrel{c}{\downarrow }}{+6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]
[tex]\left(-\cfrac{ 40}{2(-16)}~~~~ ,~~~~ 6-\cfrac{ (40)^2}{4(-16)}\right) \implies \left( - \cfrac{ 40 }{ -32 }~~,~~6 - \cfrac{ 1600 }{ -64 } \right) \\\\\\ \left( \cfrac{ -5 }{ -4 } ~~~~ ,~~~~ 6 +25 \right)\implies {\Large \begin{array}{llll} \stackrel{ ~~ seconds~\hfill feet ~~ }{\left( ~~ 1\frac{1}{4}~~ ~~ , ~~ ~~31 ~~ \right)} \end{array}}[/tex]
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