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.
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)