1. The problem statement, all variables and given/known data A model rocket is launched at 50m/s, 35 degrees above horizontal. What is the horizontal displacement when its velocity vector is at 25 degrees. 2. Relevant equations Kinematic equations, the relationship between sin cos and tan. 3. The attempt at a solution tan(25)= .4663. I interpret this as the y (sin) velocity being .4663 the x velocity. Vy = .4663Vx Vy = 50sin(25)-9.8t .4663Vx = 50sin(25)-9.8t Vx = (50sin25-9.8t)/.4663 Vx also equals 50*cos25 therefore... 50cos25 = (50sin25-9.8t)/.4663 (.4663*50cos25-50sin25)/-9.8 = t t = 3.54E^-5. Something about this answer seems very very wrong, especially when plugged back in to the x displacement equation.