Question 179599
Your problem statement seems to have some anomalies in it!
The standard formula for the height (as a function of time) of an object propelled upwards from an initial height of {{{h[0]}}} with an initial velocity of {{{v[0]}}} is given by:
{{{h(t) = -(1/2)gt^2+v[0]t+h[0]}}} where g (the constant of acceleration due to gravity) is 32 ft/sec^2, so your formula should look like:
{{{h(t) = -16t^2+v[0]t+h[0]}}}
Your formula also shows two different varaibles. t and x.
So, back to the problem but starting with the correct formul, (the d in your problem will work just as well as the h(t)) we have:
{{{d = -16t^2+64t+128}}} First, we'll find the time, t, at which the height, d, is at its maximum. The graph of the quadradtic equation that we have here is a parabola that opens downward (signified by the negative first term) so its maximum will occur at the vertex of the parabola, and this is given by:
{{{t = -b/2a}}} where a = -16 and b = 64, so...
{{{t = -(64)/2(-16)}}}}
{{{t = -64/-32}}}
{{{t = 2}}}secs.
To find the maximum height, we substitute the time, t=2secs into the original equation and solve for d.
{{{d = -16(2)^2+64(2)+128}}}
{{{d = -16(4)+128+128}}}
{{{d = -64+256}}}
{{{highlight(d = 192)}}}feet. 
The rock will reach a maximum height of 192 feet.