Why doesn't this "solution" work? 1. The problem statement, all variables and given/known data A ball is to be shot from level ground toward a wall at distance x with launch angle theta. The y-component of the velocity just as it would reach the wall, as a function of that distance x is given by the equation v_y(x) = -0.5 x + 5. Find the launch angle theta. 2. Relevant equations projectile motion 3. The attempt at a solution This problem was originally shown with some graphs, but I rewrote it with an exact equation for v_y(x) instead of the graph. Hopefully it's equally clear. Henceforth I'll use v to mean v_y and a to mean a_y since I'm not interested in the x-direction for this problem. I should say up front that I've already solved this problem "the hard way" (which was actually much harder than it appears at first glance). I had what I thought was a clever idea though and tried to apply some calculus to this problem but I seem to be doing something wrong. Essentially I said that if v = -0.5x + 5, then -0.5 = dv/dx = dv/dt * dt/dx = a(t) / v(t) -0.5 = a(t) / v(t) The second step comes from the definition of the chain rule, and the third from the fact that dv/dt = a(t) and dt/dx = 1/(dx/dt) = 1 / v(t). Since we know a(t) is equal to -9.8 m/s^2, and that dv/dx is -0.5 since it's a straight line, we have -0.5 = -9.8 / v(t) v(t) = 19.6 m/s Obviously this is wrong since velocity in the y direction isn't going to be constant. I don't see any obvious reason why the chain rule doesnt apply here although that's the most obvious place for there to have been an error. Since I know velocity isn't constant, and that a = v', I could also have re-written the equation as: -0.5 v = v' This would be a differential equation with solution v(t) = Ke^(-0.5 t) + C This doesn't seem right either though, because that means acceleration is non-constant, and this is just a simple particle in free-fall (again ignoring the x-component of motion). So I guess I'm making a fundamental error, can anyone point it out? Thanks.