1. The problem statement, all variables and given/known data A hot air balloon is traveling vertically upward at a constant speed of 4.4 m/s. When it is 27 m above the ground, a package is released from the balloon. The acceleration of gravity is 9.8 m/s^2. a) After it is released, for how long is the package in the air. b) What is its speed just before impact with the ground? 2. Relevant equations x = x0 + v0t + 1/2at^2 v = v0 + at 3. The attempt at a solution a) 0 = 27 + 4.4t + 1/2(-9.8)t^2 t = 2.839 s ***this answer is wrong. Is it wrong because v0 should equal 0 and not 4.4? b) v = 4.4 - 9.8(2.839) v = -23.42 m/s ***this answer is obviously wrong because t is wrong. But, should v0 be zero in this case as well?