Posted by **Ashley** on Wednesday, September 12, 2012 at 9:49pm.

A two-stage rocket is launched straight up from rest. It accelerates at 26.4 m/s/s for 3.92 seconds during the first stage. It accelerates at 12.3 m/s/s for 4.48 more seconds during the second stage. After second stage burnout the rocket comes under the influence of gravity only. Determine its height above launch point (in meters) after 21.2 seconds from launch.

- Physics -
**Scott**, Thursday, September 13, 2012 at 12:17pm
after 1st stage:

final velocity __ Vf1 = 26.4 * 3.92

final height __ Hf1 = .5 * 26.4 * 3.92^2

after 2nd stage:

final velocity __ Vf2 = (12.3 * 4.48) + Vf1

final height __ Hf2 = (.5 * 12.3 * 4.48^2) + (Vf1 * 4.48) + Hf1

free-fall time is __ 21.2 s - 3.92 s - 4.48 s , or 12.8

H = (-.5 * g * 12.8^2) + (Vf2 * 12.8) + Hf2

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