[SOLVED] Projectile motion and time 1. The problem statement, all variables and given/known data A hot air balloon rises from the ground with a velocity of (2.00 m/s) in the y direction. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (5.00 m/s) in the x direction, relative to the balloon. When opened, the bottle is 6.00 m above the ground. How long does the cork remain in the air? 2. Relevant equations There are other parts of this problem that I've already solved. For example, I've found that the initial velocity and direction of the cork, as seen by an observer on the ground, is 5.39 m/s at 21.8 degrees above horizontal (w/ respect to the balloon) with these equations: V= [(5.0)^2+(2.0)^2]^1/2 theta= arctan (2.0/5.0) I've also found that the maximum height above the ground attained by the cork is 6.204 meters by finding: h= [(5.39^2)sin^2(21.8)]/2(9.81) and adding it to the initial height of 6.00 m. 3. The attempt at a solution I know that the equation to find the time that a projectile is in the air is as follows: t= [2Vo*sin(theta)]/g but I know that that only accounts for the time that it will take the cork to reach a height of 6 meters. What I don't know is how to account for the rest of the time.