1. The problem statement, all variables and given/known data http://img524.imageshack.us/img524/6899/tempnb3.th.png [Broken] 2. Relevant equations y = y0 - 1/2gt^2 + voy*t 3. The attempt at a solution Well there are two methods that I used for solving this, but I think only one is correct: First, the correct method: 1) Find the time at the top of the parabola and use the distance from the hoop to determine the time. 12 = 5 - 16t^2 + v0*t*(5/13) x = v0*(12/13)*t = 24 t = .433s v0 is 60 ft/s 2) Find the time at y = 3 (to hit the second slope) 0 = 5-3 - 16t^2 + vo(5/13)*t t = 1.524s 3) Find h x = vo(12/13)*t = h + 24 --> h = 60.4 ft I think everything I did is correct, however, my TA in lecture today said that the v0 should be between 50 ft/s - 60 ft/s and h should be between 40 ft and 47 ft. However, he also said that he didn't quite know how to get the correct solution.... (he was looking at the solutions manual) I get 60ft/s but my h is 60.4 ft however, if I use the 2nd method I get within the range, but isn't my method correct and the second method wrong? method 2: 12 = 5 - 16t^2 + v0*t*(5/13) v = vo(5/13) - gt v = 0 since it's at the top of it's path v0 = gt * 13/5 v = 55 ft/s ... t = .661 s But isn't that wrong because if I plug in .661 s into x = vo(12/13)*t I get ~33ft, whereas it should be 24 ft at the top of the path.. Can anyone advise me on which method is correct?