The first step to answering the Problem is to know the formulas you would need. This is a Constant-Acceleration Problem and there are 4 fundamental equations or recipes you would need if ever you are face with a similar problem in the future. Here are the formulas:

1.) To look for the velocity: v=vi + at; where (v) is the velocity, (vi) is the initial velocity and (a)(t) is acceleration multiplied to time.

2.) To look for the displacement in terms of velocity: d-di = (1/2)(vi+v)(t); where d is the displacement and (di) is the initial displacement.

3.) To look for displacement in terms of a: d-di=(vi)(t) + (1/2)(a)(t^2); where (t^2) is time squared (these equations are REALLY easy to understand if you analyze them spatially.)

and last:

4.) the velocity in terms of a and d: (v^2)=(vi^2) + 2a(d-di)

For the problem you have above, you need the third and first equation.

Separate the problem into 2 instances: distance traveled with fuel and distance traveled without fuel then add them together and you will get the answer: 931.41552 meters.

First Part: With Fuel

Use equation number 3 to get d:

Since we know a, t , and that vi and di are equal to zero equation 3 is reduced to:

d=(1/2)(29.4m/s^2)([3.98s]^2)

d=232.85388

Second Part: Without Fuel

First we use equation1 to know the velocity acquired with fuel and since we know vi is zero, and we know a and t:

v=(29.4)(3.98)

v=117.012

Next we look for the time it takes for this velocity to reach zero, since there is no more fuel we assume gravity will pull the ship back. We will again use equation1 but this time to look for the time. Our goal is for v=0, we set vi=117.012, and acceleration due to gravity a=-9.8. Equation 1 becomes:

0=(117.012)-(9.8)(t)

(9.8)(t)=117.012

t=(117.012)/(9.8)

t=11.94 seconds before v=0

Last we use equation3 to know the displacement and we know di=232.85388, previously we know vi=117.012, t=11.94, and acceleration due to gravity a=-9.8. Equation3 becomes:

d-232.85388=(117.012)(11.94)-(1/2)(9.8)([11.94]^2)

d-232.85388=698.56164

d=931.41552

d=931

v = at = 29.4 x 3.98 = 117.012 m/s

and has also gained a height of (using equation above) of:

distance s= 232.85 m

after this gravity causes a constant deceleration of about 9.8 m/s^2

v^2 = u^2 + 2as

where v is end velocity is 0 m/s^2

0 = u^2 + 2as

s = -u^2/2a

note gravity is decceleration = -9.8 m/s^2

s = 117^2/2×9.8 = 698.4 m

adding distances

dist total = 232.9 + 698.4 = 931.3 m

u = 0 (it starts from rest)

t = 3.98, a = 29.4 (given)

Then when the fuel runs out it has travelled 232.85 m