Question 359046
No, that's not correct.
You're solving for when {{{h=0}}}.
{{{h(t)=56t-16t^2}}}
{{{h(t)=8t(7-2t)}}}
{{{h=0}}} when 
{{{t=0}}} and {{{7-2t=0}}}
{{{t=0}}} and {{{2t=7}}}
{{{t=0}}} and {{{t=7/2}}}
Using a symmetry argument, then the maximum height must occur when time is between these two values, {{{t=(1/2)(7/2)=7/4}}}
Find {{{h(7/4)}}}
{{{h(7/4)=56(7/4)-16(7/4)^2}}}
{{{h(7/4)=98-49}}}
{{{highlight(h(7/4)=49)}}}ft
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You could also have converted the height equation to vertex form, {{{y=a(t-h)^2+k}}} since the maximum value occurs at the vertex.
I use {{{y}}} instead of {{{h(t)}}} here because ({{{h}}},{{{k}}}) is traditionally the vertex.
Complete the square.
{{{y=-16t^2+56t}}}
{{{y=-16(t^2-(7/2)t)}}}
{{{y=-16(t^2-(7/2)t+49/16)+16*(49/16)}}}
{{{y=-16(t-(7/4))^2+49}}}
So then the maximum value of {{{y=49}}} occurs at {{{t=7/4}}}
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{{{drawing(300,300,-2,4,-5,90,circle(0,0,0.15),circle(7/2,0,0.15),grid(1),circle(7/4,49,0.15),blue(line(7/4,-10,7/4,100)),graph(300,300,-2,4,-5,90,56x-16x^2))}}}