Amissile is launched upward with an acceleration of 100m/s^2. eight seconds later, the engine shuts off.

what height does the missile achieve?
what is the total time it is in flight?
what speed does it have when it strikes the ground?
what is the greatest speed of the missile?

1.) The height of the missile should include when the engine was still working and when it turned off. The moment it turned off, only gravitational force is working on it.

H = height during the 8-second launch + height of upward gravitational motion
H = h + y
H = (v₀t + 0.5at²) + (v₀²/2g)
But first, let's calculate v₀ from the acceleration equation:
a = (v - v₀)/t
Since for sure the missile started from rest, v₀ = 0.
100 = (v - 0)/8 s
v = 800 m/s
Note, that v₀ for h and v₀ for y are not the same. For h, v₀ = 0 because the missile started from rest. For y, v₀ = v = 800 m/s, because at this point, the engine turned off. Thus,

H = ((0)(8 s) + 0.5(100 m/s²)(8 s)²) + ((800 m/s)²/2(9.81 m/s²))
H = 3,200 m + 32,619.78 m
H = 35,819.78 meters

2.) The total time of flight is 8 seconds plus the time for height y. 

Total time = 8 s + √2y/g
Total Time = 8 s + √2(32,619.78 m)/(9.81 m/s²)
Total time = 89.55 s

3.) When it strikes the ground, the velocity would be zero because it has stopped.

4.) The greatest speed is just before the missile strikes the ground. This is called the velocity at impact having the equation,

v = √2gH = √2(9.81 m/s²)(35,819.78 m)
v = 838.32 m/s