A spaceship lifts off from the Earth with constant acceleration. 3.4 seconds int

Question

A spaceship lifts off from the Earth with constant acceleration. 3.4 seconds into the flight, a small piece of the ship breaks off and falls back to the Earth. If the piece hit the ground 6.6 seconds after breaking off from the spaceship, what is the spaceship’s acceleration (in m/s^2)?

I think the answer should be around 6m/s^2. My attempt:

A = acceleration of rocket
V = velocity
S = distance
Phase 1:   (from takeoff to when piece falls)
Height = 5.78*A
V1 = 3.4*A Phase 2: (from when piece falls to when Velocity = 0)
?S = 0.589*(A^2)
?t = 0.346*(A) So the height when Velocity = 0 is (5.78*A + 0.346*A^2)
Now what?
I think the answer should be around 6m/s^2. My attempt:

A = acceleration of rocket
V = velocity
S = distance
Phase 1:   (from takeoff to when piece falls)
Height = 5.78*A
V1 = 3.4*A Phase 2: (from when piece falls to when Velocity = 0)
?S = 0.589*(A^2)
?t = 0.346*(A) So the height when Velocity = 0 is (5.78*A + 0.346*A^2)
Now what?

Explanation / Answer

H1: (1/2)A*3.4^2 = 5.78*A, yes. V1 = 3.4A, yes. You can go right from your phase 1 into vertical projectile motion. The piece has initial upward velocity V1, the height H1, and downward acceleration equal to gravity. in 6.6 seconds it peaks and goes down H1 distance from where it started so: V0*t + (1/2)*A*t^2 = change in position V1*t - (1/2)*G*t^2 = -H1 where t=6.6... substitute everything in terms of A and solve: 3.4*A*6.6 - (1/2)*9.81*6.6^2 = -5.78A A = 7.5713 m/s^2

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