Question 182343
1.  I hope you know derivatives, because the easiest way to solve this is to take the first derivative, and you know at the top it will have a slope of 0, we can solve for the time there, and then substitute that back in to get the height.

h'(t)=-32t+256=0 --> -32t = -256 --> t = 8 (sec)

h(8) = -16(8^2)+256*8 = 1024 (ft)

2.  This isn't as hard, we know that it rests at h(t) = 0.  Note, this will have two roots, first at t = 0, since the rocket starts from the ground.

h(t) = 0 = -16*t^2 + 256t -->  t(-16t+256) = 0

Again, one root is t = 0, other is -16t + 256 = 0
16t = 256 --> t = 16 (sec)