One-D Kinematic Question w/ rising hot air balloon

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SUMMARY

The problem involves a hot-air balloon rising at a constant rate of 2.3 m/s and a camera tossed upward with an initial speed of 14 m/s from a position 2.5 m below the balloon. The key equations used are the constant linear acceleration equations: v = v(i) + a*t and x = x(i) + v(i)t + 1/2at^2. The balloon's acceleration is zero since it is moving at a constant speed, while the camera experiences an acceleration of -9.8 m/s² due to gravity. The solution requires equating the positions of the balloon and the camera to find the height at which they meet.

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millergc
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Homework Statement



A hot-air balloon has just lifted off and is rising at a constant rate of 2.3m/s. Suddenly, one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 14m/s.

If the passenger is 2.5m above her friend when the camera is tossed, how high is she when the camera reaches her.


Homework Equations



Constant linear acceleration equations

v=v(i) +a*t
x=x(i) + v(i)t + 1/2at^2


The Attempt at a Solution



The part of this problem that confuses me is the acceleration of the camera. I know that the acceleration of the balloon is just -9.8 but how can I find the acceleration of the camera without time, etc. I tried solving where x(balloon)= x(camera) but i used -9.8 as the acceleration for both and i do not believe i got the right time.

Any help would be great.
Thanks.
 
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millergc said:
The part of this problem that confuses me is the acceleration of the camera.
The acceleration of the camera is the same as any other body in free fall.
I know that the acceleration of the balloon is just -9.8
No. You are told that the balloon is rising at a constant rate--so what must be its acceleration?
 
Doc Al said:
The acceleration of the camera is the same as any other body in free fall.

No. You are told that the balloon is rising at a constant rate--so what must be its acceleration?

zero, ok i understand, i just thought -g would still act on the balloon but it wouldn't

Thanks
 

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