Question 847122
Analyze the quadratic formula a little
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
You can break this up into 2 parts, both of which are true:
{{{ t[1] = (-b + sqrt( b^2 - 4*a*c )) / (2*a) }}}
and
{{{ t[2] = (-b - sqrt( b^2 - 4*a*c )) / (2*a) }}}
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Now show it this way:
{{{ t[1] = -b/(2a) + sqrt( b^2 - 4*a*c ) / (2*a) }}}
and
{{{ t[2] = -b/(2a) - sqrt( b^2 - 4*a*c ) / (2*a) }}}
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What these last 2 equations are saying is that one
of the roots is this far {{{ sqrt( b^2 - 4*a*c ) / (2*a) }}}
in one direction, and the other root is this far
{{{ sqrt( b^2 - 4*a*c ) / (2*a) }}} in the opposite direction
with {{{ -b/(2a) }}} right in the middle.
So, {{{ -b/(2a) }}} is the t-coordinate of the max or the min 
( exactly 1/2 way between the roots )
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With your problem:
{{{ s(t) = -16t^2 + 48t }}}
{{{ a = -16 }}}
{{{ b = 48 }}}
{{{ t(max) = -b/(2a) }}}
{{{ t(max) = -48/( 2*(-16)) }}}
{{{ t(max) = 48/32 }}}
{{{ t(max) = 1.5 }}} 
Now plug this back into your equation
to get {{{ s(max) }}}
{{{ s(max) = -16*1.5^2 + 48*1.5 }}}
{{{ s(max) = -16*2.25 + 72 }}}
{{{ s(max) = -36 + 72 }}}
{{{ s(max) = 36 }}} ft
Here's the plot:
{{{ graph( 400, 400, -1, 4, -5, 40, -16x^2 + 48x ) }}}