Free fall with horizontal velocity

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.
 

LowlyPion

Homework Helper
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Don't you want instantaneous velocity?

After 2 seconds won't the dropping V be -19.6 m/s

After 3 horizontal will be 3 and vertical -g*t = -29.4 m/s

And overall |V| will be (32 +29.42)1/2 and direction via arc tan?
 
220
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Don't you want instantaneous velocity?

After 2 seconds won't the dropping V be -19.6 m/s

After 3 horizontal will be 3 and vertical -g*t = -29.4 m/s
I know I'm nit-picking, but doesn't the problem state that the spacecraft is falling in the y direction. So those speed should be positive.
 

LowlyPion

Homework Helper
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It doesn't really matter so long as the direction chosen as positive is maintained through out.

I merely continued to use the direction already indicated by the OP's equation for y.

In general my preference is usually positive up anyway.
 
nah, looking for displacement, and your answer doesn't match their's anyway. They still get a total displacement of 19.6 meters to t=2. But looking at it more, they ignore the acceleration of the object over the first two seconds when calculating displacement over (2,3)
 

LowlyPion

Homework Helper
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4
nah, looking for displacement, and your answer doesn't match their's anyway. They still get a total displacement of 19.6 meters to t=2. But looking at it more, they ignore the acceleration of the object over the first two seconds when calculating displacement over (2,3)
Are you looking for displacement or velocity? The original problem says velocity doesn't it?

As to my equation I see that I took the horizontal V to be 3 and in rereading the problem I see that should have been 6m/s at 3s.
 
doh... i miscopied the problem... that should say displacement after 2 seconds, sorry about that. Anyway, I talked to the professor, it was an answer book error, thanks for trying anyway.
 

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