A ball is thrown up onto a roof, landing 3.70s later at height h = 21.0m above the release level. The ball's path just before landing is angled at θ = 54.0˚ with the roof. Find the horizontal distance it travels. Answer: 33.5m Related Equations: Xf = Xo + 1/2(Vf + Vo)t _____________________ My approach on this problem was to find the velocity the moment before impact with the roof and to use basic trigonometry to calculate the horizontal velocity of the ball. From the given facts I can infer that: Xf = 21m Xo = 0m t = 3.70s Vo = ? Vf = Vo*sin(54) Now, plugging this into the equation above I get: 21 = 1/2(Vo*sin(54) + Vo)3.70 42 = (Vo*sin(54) + Vo)3.70 11.35 = .81Vo + Vo 11.35 = 1.81Vo 6.27 = Vo With Vo I can calculate the horizontal velocity which would be: Vx = Vo*cos(54) Vx = 6.27(.59) Vx = 3.69 And the distance should therefore be: d = VxT d = 3.69(3.70) d = 13.65 However, the answer is not 13.65m and after scratching my head and approaching this problem through multiple, yet similar, angles I finally decided to ask for help. Can somebody help me and point out what I did wrong? I thought I had the right idea but I can't seem to get the right answer? Thank you for taking the time to read my question. Sincerely David A.P.