How High is the Rocket When the Canister Hits the Launch Pad?

[blite777]

Homework Statement

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]

Homework Equations

V^x=V_0x+a_xt
x=x₀+V_0xt+1/2 a_xt²
V_x²=V_0x²+2a_x(x-x₀)

The Attempt at a Solution

235=0+0+1/2 (3.30)t²
235=1.65t²
t²=235/1.65
t=11.93s

V_x=0+9.8(11.93)
V_x=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=932.94mI 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

 

[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.