Projectile Motion - V0 is 5 times v @ max height?!? Thanks prior, all... 1. The problem statement, all variables and given/known data A projectile's launch speed is 5 times its speed at maximum heigth. Find launch angle theta0. 2. Relevant equations Not sure which are relevant on this one, but the given for the section are: x - x0 = (v0 cos theta0)t, y - y0 = (v0 sin theta0)t - 1/2gt2, vy = v0 sin theta0 - gt, vy2 = (v0 sin theta0)2 - 2g(y - y0) 3. The attempt at a solution So, I've drawn the path, defining the maximum heigth as vm. At vm, the vertical component, vmy, must be 0. vmx must be constant and would equal the same as it started, vmx = vox In the problem, the initial velocity, v0 is 5 times vm, so v0 = 5(vm) = 5(vox) Not sure where to go from here, I tried to break vo into its components, but just ended up with voy2 = 24vox2 Any ideas? Thanks again!