A hot-air balloon has just lifted off and is rising at the constant rate of 1.8 . Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward. If the passenger is 3.1 above her friend when the camera is tossed, what is the minimum initial speed of the camera if it is to just reach the passenger? (Hint: When the camera is thrown with its minimum speed, its speed on reaching the passenger is the same as the speed of the passenger.) I tried to use V^2=Vi^2 + 2aΔX to solve for Vi I plugged in (1.8)^2=Vi^2 +2(-9.8)3.1 and got Vi=8m/s but that answer is wrong. Where did I go wrong?