# Rocket Launching

1. Apr 9, 2010

### blite777

1. The problem statement, all variables and given/known data
You are working as an engineer in a space company. During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.30m/s2. When it is 235m above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity. You can ignore air resistance. How high is the rocket when the canister hits the launching pad, assuming that the rocket does not change its acceleration?
[The back of book Answer is: 945m]

2. Relevant equations
Vx=V0x+axt
x=x0+V0xt+1/2 axt2
Vx2=V0x2+2ax(x-x0)

3. The attempt at a solution
235=0+0+1/2 (3.30)t2
235=1.65t2
t2=235/1.65
t=11.93s

Vx=0+9.8(11.93)
Vx=116.96m/s

116.962=02+2(9.8)(x-235)
116.962=2(9.8)(x-235)
116.962/2(9.8)=x-235
116.962/2(9.8) + 235 = x
x=932.94m

I don't know to get 945m =( .. I would appreciate it if someone can solve this problem and show the steps correctly..
Thank you so much in advance

2. Apr 11, 2010

### phyzguy

You did the first step right, where you calculated that it takes 11.93 s for the rocket to reach 235 m. However, your second step is all wrong. You want to calculate how long after the separation it takes for the discarded fuel canister to reach the ground. Use you second equation, with x=0, x0 = 235m, v0 =(3.3 m/s^2 * 11.93s), and a0=-9.81 m/s^2. Then plug this time delay into the second equation to see how much higher the rocket goes during this time. When I did this, I got 945m.