How high does a rocket go with a constant acceleration of 25m/s^2?

[lolbits]

Homework Statement

A rocket is sent straight up in the air from the ground. It travels with a constant acceleration of $$25m/s^2$$. It will have a constant acceleration as long as the engine gets fuel. After 20 seconds, the engine suddenly shuts off. Calculate how high the rocket will get. (No air friction).

Homework Equations

$$v^2 = v_0^2 + 2 a \Delta x$$
$$x = x_0 + v_0 t + (1/2) a t^2$$

The Attempt at a Solution

I first drew a picture of the situation. Then, I calculated the distance traveled with engine power.

Gives me:
d(after 20s)=(1/2)*25*20^2 = 5000 m.
Velocity at 20 s = sqrt(2*a*d) = 500 m/s.

d(after no engine power) = (v^2=(v_0)^2+2ad)
Want to solve for d, so I rearrange and get: (v^2-(v_0)^2)/(2*(-9.81)
I get that d=12742m.
12742m + 5000m = 17742m = 17.742km.

The book says that the answer should be 18 km.
Have I done anything wrong?

 

[rl.bhat]

No. You are right. But check the significant numbers.

 

[OpticalSuit]

you are right just round up