Hello all, I've been having some problems regarding the fall of an object in the vertical direction that has air drag which varies quadratically with speed. The problem is as shown: A gun is fired straight up. Assuming that the air drag on the bullet varies quadratically with speed. Show that speed varies with height according to the equations v^2 = Ae^(-2kx) - g/k (upward motion) V^2 = g/k - Be^(2kx) (downward motion) in which A and B are constants of integration, g is the acceleration of gravity, k=cm, where c is the drag constant and m is the mass of the bullet. (Note: x is measured positive upward, and the gravitational force is assumed to be constant.) I've started on the question, but I keep coming to the wrong answer. I shall type out what I have done. The general equation would be: (downward) m dv/dt = mg - cv^2 = mg(1 - cv^2/mg) .'. dv/dt = g(1 - v^2/v(sub t)^2) where v(sub t) is the terminal speed, sqrt(mg/c) We now play around with the differential equation to obtain the independant variable for distance rather than time. dv/dt = dv/dx dx/dt = 1/2 dv^2/dx So now we can write the equation as: dv^2/dx = 2g(1 - v^2/v(sub t)^2) and we solve the equation using substitution method u= 1 - v^2/v(sub t)^2 so that du/dx = -u(2g/v(sub t)^2) u=u(naught)e^[-2gx/[v(sub t)^2]] and we know u(naught) = 1 - v(naught)^2/v(sub t)^2 putting everything together, we get (using k= c/m, v(sub t)= sqrt(mg/c) v^2 = g/k(1-exp^[-2kx]) + v(naught)^2.exp^[-2kx] Likewise for upward, we get: v^2 = g/k(1+exp^[2kx]) + v(naught)^2.exp^[2kx] I don't seem to be able to continue any further to reach the proof. Could someone profound in this area give me some assistance? Thanks in advance, Levi.