Question 688908
You can relate 2 variables- {{{ h }}} plotted on
the vertical axis, and {{{ d }}} plotted on the horizontal 
axis. I know that the 2 roots of the equation are:
{{{ d = 0 }}}
{{{ d = 1600 }}}
This is because {{{ d = 0 }}} is where the cannonball is
fired from and {{{ d = 3200 }}} is where it lands.
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The general form for a parabola is
{{{ h = a*d^2 + b*d + c }}}
You are given 2 points, ( 0,0 ) and ( 1600, 0 )
(0,0)
{{{ 0 = a*0^2 + b*0 + c }}}
{{{ 0 = c }}}
So now the equation looks like
{{{ h = a*d^2 + b*d }}}
-------------------
(1600,0)
{{{ 0 = a*(1600)^2 + b*1600 }}}
{{{ a*(1600)^2 = -b*1600 }}}
{{{ 1600a = -b }}}
{{{ b = -1600a }}}
----------------
The vertex is at {{{ -b/(2a), h[max] }}}
{{{ -b/(2a) = 1600a/(2a) }}}
{{{ -b/(2a) = 800 }}}
You are given that {{{ h[max] = 3200 }}}
{{{ 3200 = a*800^2 + b*800 }}}
{{{ 4 = 800a + b }}}
Since {{{ b = -1600a }}}
{{{ 4 = 800a - 1600a }}}
{{{ 4 = -800a }}}
{{{ a = -1/200 }}}
and
{{{ b = -1600a }}}
{{{ b = -1600*(-1/200 ) }}}
{{{ b = 8 }}}
The equation is:
{{{ h = -(1/200)*d^2 + 8d }}}
------------------------
check answer:
Does it go through (0,0)?
{{{ h = -(1/200)*d^2 + 8d }}}
{{{ 0 = -(1/200)*0^2 + 8*0 }}}
{{{ 0 - 0 }}}
OK
Does it go through (1600,0)?
{{{ h = -(1/200)*d^2 + 8d }}}
{{{ 0 = -(1/200)*1600^2 + 8*1600 }}}
{{{ 0 = -2560000/200 + 12800 }}}
{{{  0 = -12800 +12800 }}}
{{{ 0 = 0 }}}
OK
Is the vertex at ( 800, 3200 )?
{{{ h = -(1/200)*d^2 + 8d }}}
{{{ 3200 = -(1/200)*800^2 + 8*800 }}}
{{{ 4 = -(1/200)*800 + 8 }}}
{{{ (1/200)*800 = 4 }}}
{{{ 4 = 4 }}}
OK
Here's a plot of the parabola.

{{{ graph( 600, 600, -500, 1800, -500, 3500, -(1/200)*x^2 + 8x ) }}}