The question: An object has a Velocity V1 and an angle A (Alpha) initially. Later in its path, this same object has an Angle B (Beta) find the time elapsed from point A to point B Okay so i have an answer.. i just need some one to tell me if what i'm doing is correct./answer is correct. http://postimage.org/image/xsei89eg1/ [Broken] here is an image of the problem. Okay so.. V1x = V1cosA V1y = V1sinA Let V2 be the variable for the second velocity. Since horizontal component of velocity is constant through projectile motion.. V1x=V2x V2x = v1cosA Through trigonemtry we can resolve the Y component for V2, which i need for further calculation TanB = V2y/V2x V2y= V2xTanB Now i have all i need and i can use uniform accelerated formula, to solve for The time... Vf=Vi + gt V2y = V1y + gt (v2y - v1y)/g = t Answer: t = (v1cosATanB - V1sinA)/g or simplifieed too... V1(cosA*TanB - SinA)/g (m/s)/(m/s^2) = s (seconds) Units match.