Kinematics Problem on hot air balloon

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The hot air balloon rises at a constant rate of 1.30 m/s, while a passenger's friend tosses a camera upward with an initial speed of 9.40 m/s from a point 2.30 m below her. To solve the problem, one must consider the distances traveled by both the balloon and the camera over the same time interval. The balloon's distance can be expressed as x, while the camera's distance is x + 2.30 m. By applying kinematic equations, the time it takes for the camera to reach the passenger and her height at that moment can be determined.
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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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