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?