[tex]The\ axis\ of\ symetry\ of\ the\ graph\ quadratic\ function:x=h\\\\where\ "h"\ is\ from\ vertex\ form\ of\ quadratic\ function:\\\\f(x)=a(x-h)^2+k\\----------------\\\\f(x)=ax^2+bx+c\ then\ h=\frac{-b}{2a}\\----------------\\\\f(x)=-16x^2+256x\to a=-16;\ b=256;\ c=0\\\\h=\frac{-256}{2(-16)}=\frac{-256}{-32}=\boxed8\to\boxed{\boxed{x=8}}[/tex]
[tex]When\ it\ to\ fall\ back\ to\ the\ ground?\\In\ zeros\ of\ the\ function! ;)\\\\f(x)=-16x^2+256x=0\\\\-16x^2+256x=0\ \ \ \ |divide\ both\ sides\ by\ (-16)\\\\x^2-16x=0\\\\x(x-16)=0\iff x=0\ or\ x-16=0\\\\\boxed{x=16}\\\\Answer:\boxed{\boxed{C}}[/tex]
