# Help with max rocket height. Heights differ with methods.

1. Sep 2, 2015

1. The problem statement, all variables and given/known data
A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 30 m/s^2 for 30s, then runs out of fuel. Ignore any air resistance effects. What is the rocket’s maximum altitude?

2. Relevant equations
Δx=V0t+(1/2)at2
V=V0+at
2aΔx=V2-V02

3. The attempt at a solution
Δx=(1/2)(30m/s2)(30s)2=13500m

V=0(m/s)+(1/2)(30m/s2)(30s)=900m/s
2(-9.81m/s2)Δx=02-9002m/s
=> Δx=41300m

Total height: 55000m

However, I did it another way and received a different result.
V=V0+at
0=900+(-9.81)t
t=91.7s

Δx=(900m/s)(91.7s)=82530m
Total height: 96000m

Why is this?

2. Sep 2, 2015

### Titan97

In your second method, there is a missing $-\frac{1}{2}gt^2$ term. After it runs out of fuel, its acceleration will be g but not 0.