Question 144914
{{{h = 64t - 16t^2}}}
If you substitute {{{t=0}}}, you see that {{{h=0}}}
That is the time and the height when the object is
released. 
What you want to find is another time when {{{h=0}}},
and that will be the time when the object has come
back to earth, so
{{{64t - 16t^2 = 0}}}
{{{t(64 - 16t) = 0}}}
This equation is true if either of the factors, {{{t}}}, or 
{{{64 - 16t}}} is zero
{{{64 - 16t = 0}}}
{{{t = 4}}}
So the object is in the air 4 sec before it hits the ground