1. The problem statement, all variables and given/known data There is a hot air balloon on the ground. It moves upwards at a constant acceleration. After 30 seconds someone throws a ball horizontally. An observer times the ball in the air to be 10 seconds. Knowns: Time of the ball: 10seconds Gravity: 9.8m/s You are to find: a) the balloons acceleration b) initial height at which the ball is thrown c) height of the baloon when the ball hits the ground d) ask God for forgiveness Answers a) 0.65m/s2 b) 294m c) 522m 2. Relevant equations vf^2=vi^2+2ad d=vi(t)+0.5(a)t^2 d=vf(t)+0.5(a)t^2 d=vt vf=vi+a(t) d=0.5(vi+vf)t 3. The attempt at a solution I tried to combine formulas but when I did i had gotten incorrect answers. There are not enough knowns to solve it directly. The class (20 people) couldn't get it after 30min.. This is Grade 12 Physics.. I know that because the ball is moving horizontally then the velocity going horizontal is constant, and that because as it is thrown the balloon is moving upwards then the ball also has a velocity upwards, creating a parabola.