s = -16t^2 + 560t
b) By estimate I assume they mean graph it and guess where the highest point is. So graph it.
c) Analytically we know that v = 0 ft/s at maximum height. If you don't have Calculus at your disposal you probably know that v = v0 + at, so v = 560 - 32t. Set v = 0 ft/s and solve for t.
If you do know Calculus then v = ds/dt = 560 - 32t again. Either way t = 17.5 s.
d) v = 560 - 32t. Just plug in the times. I get v = 240 ft/s and v = -240 ft/s respectively.
e) The speed of the rocket is different from the velocity in one respect: speed does not have direction. The speed is the magnitude of the velocity so the answer is |v| = 240 ft/s in both cases.
f) When is the rocket on the ground? When s = 0 ft. Thus you need to solve the equation 0 = -16t^2 + 560t This will give you your t value, then use v = 560 - 32t. Alternately we have the equation
v^2 = v0^2 + 2a(s - s0)
We know that s = s0 = 0 ft here, so this means
v^2 = v0^2
v = (+/-)v0
Since all we care about is the speed, which is the magnitude of the velocity we have that:
|v| = |v0| = 560 ft/s, which had better be what you get using the first method as well.