One-D Kinematic Question w/ rising hot air balloon

In summary, the conversation discusses a problem involving a hot-air balloon rising at a constant rate and a passenger who realizes she has left her camera on the ground. The camera is tossed upward by a friend and the question is asked about how high the passenger will be when the camera reaches her. The relevant equations and confusion about the acceleration of the camera are also mentioned. After clarification, it is determined that the acceleration of the camera is the same as any other body in free fall and the balloon's acceleration is zero since it is rising at a constant rate.
  • #1
millergc
2
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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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  • #2
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?
 
  • #3
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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