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.
Constant linear acceleration equations
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.