Kinematics Problem on hot air balloon

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SUMMARY

The discussion centers on a kinematics problem involving a hot air balloon rising at a constant rate of 1.30 m/s and a camera tossed upward with an initial speed of 9.40 m/s. The passenger is initially 2.30 m above the friend who throws the camera. By applying kinematic equations, the distances traveled by both the passenger and the camera can be equated to determine the time it takes for the camera to reach the passenger and the height of the passenger at that moment. The solution involves solving for time (t) and the distance (x) traveled by the passenger.

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A hot air balloon has just lifted off and is rising at the constant rate of 1.30 m/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 9.40 m/s. If the passenger is 2.30 m above her friend when the camera is tossed, how much time does it take for the camera to reach her? How high is the passenger when the camera reaches her?

I am not completely sure how to even start this. I know you have to factor in that the air balloon is rising as she is throwing the camera in the air, but i do not know how to factor that inforation in. Help would be greatly appreciated. Thanks
 
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Hot air balloon is moving with constant acceleration. Camera is tossed with some initial velocity. When the passenger catches the camera, the distance traveled by her is x and the distance traveled by camera is x + 2.3 m. These distances were traveled in the same time interval. Using kinematic equations, find t and x.
 

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