Arocket moves upward, starting from rest with an acceleration of 29.4 m/s^2 for 3.98 s. it runs out of fuel at the end of 3.98 but does not stop. how high does it rise above the ground

The distance d₁ it rises from rest while the engine is burning is given by
d₁ = d₀ + v₀t + (1/2)at²
d₁ = 0 + 0 + (1/2)·(29.4 m/s²)·(3.98 s)² = 232.85 m
So it gets to 232.85 m and then runs out of fuel.  Its velocity v₁ at this point is given by
v₁ = v₀ + at = (29.4 m/s²)·(3.98 s) = 117 m/s
At this point, gravity begins to slow it down until it reaches its peak where its velocity v₂ is zero.
v₂² = v₁² + 2ad₂
where d₂ is the distance it rises until v=0
Since gravity is decelerating the rocket, a = -g, and we have
0² = (117 m/s)² + 2(-9.8 m/s²)d₂
0 = (117)² - (19.6)·d₂
0 = 13,689 - (19.6)·d₂
d₂ = 13,689/19.6 = 698.42 m
So the total height it rises is given by
d₁ + d₂ = 232.85 m + 698.42 m
= 931.27 m

When is this

Work is reduced while time is decreased