1. The problem statement, all variables and given/known data A ball is thrown at 60 degrees above x-axis. It remains in the air for 2.9 seconds. What is the final displacement in the x (range) if the ball was thrown with the same initial velocity, but at an angle of 45 degrees above x-axis instead? 2. Relevant equations y = (Voy * t) + (1/2 * ay * t^2) Voy = Vo * sin(theta) Vox = Vo * cos(theta) x = 1/2(Vox + Vx)t 3. The attempt at a solution So plugging into the first equation I get: y = (Voy * t) + (1/2 * ay * t^2) 0 = (Voy * 2.9) + (1/2 * -9.8 * 2.9^2) Voy = 14.21 Taking the result into the second equation I get the initial velocity: Voy = Vo * sin(theta) 14.21 = Vo * sin(60) Vo = 16.41 Am I on the right track so far? From here I would think I'd take it into the first equation again with the 45 degree angle so: y = (Voy * t) + (1/2 * ay * t^2) 0 = (16.41sin(45)*t) + (1/2 * -9.8 * t^2) 0=11.60t-4.9t^2 t = 2.37 So I've got the time, and Vox from Vox = Vo * cos(theta) since Vo is the same for both throws. For the equation x = 1/2(Vox + Vx)t (substitute Vox = Vo * cos(theta)) x = 1/2(16.41cos(45) + Vx)2.37 All I'm missing is Vx, I am lost where to go from here.