(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An airplane releases a spacecraft, after 2 seconds the spacecrafts engine starts providing it with a horizontal velocity of 6m/s relative to the aircraft. What is the magnitude and direction of the velocity after 2 seconds? After 3? Assume the spacecraft is in free-fall in the y direction.

2. Relevant equations

x=vit+1/2at^2

y=-1/2gt^2

3. The attempt at a solution

It seemed pretty easy... Plug t=2 into the y equation and get a y displacement of 19.6 meters. Then plug in t=3 and arrive at a y displacement of 44.2m, and an x displacement of 3m. Then take the arctan(44.19/3) to get an angle of -86 degrees. The problem is that the answer book appears to split the reference frame at t=2. They say that the y displacement between 2 and 3 seconds is only 4.91m. This doesn't make sense because the object should continue to accelerate downward regardless of the change in x acceleration. I would just assume that this was an error in the answer book, but that would be 3 in a row, and I don't trust myself that much. Any help would be appreciated.

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# Homework Help: Free fall with horizontal velocity

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