Question 195469
{{{h=32 +48t -16t^2}}}
At what time is the height 64?
{{{64 = 32 +48t -16t^2}}}
{{{0 = -16t^2 + 48t - 32}}}
{{{0 = t^2 -3t + 2}}}
{{{0 = (t-2)(t-1)}}}
t = 1 and t =2

It is at 64 feet in two places. First at 1 second in, then again at 2 seconds.
Here is the graph of the arc. Note that the part of the graph to left of time=0, does not apply.
{{{graph(400,400,-1,4,-10,90,-16x^2 + 48x + 32)}}}