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Vertical fall with quadratic air drag. 
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#1
Aug706, 01:56 AM

P: 20

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(1exp^[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. 


#2
Aug706, 04:59 PM

Mentor
P: 41,085

Maybe check the sign of the air drag in your first equation. The force from the drag is in the same direction as the gravitational force on the way up.



#3
Aug706, 05:55 PM

Mentor
P: 6,246

Let me expand a bit on berkeman's comments.
You have used v = dx/dt in dv/dt = dv/dx dx/dt = 1/2 dv^2/dx. Since x is a vector so is v, i.e., v is not speed. In other words, x, v =dx/dt, and a = dv/dt are all vectors that have their directions given by their signs. Consequently, for the downward case, the starting equation should be m dv/dt = mg + cv^2. Gravity acts downward, drag acts upward, the direction of the net force m dv/dt is given by the sign of dv/dt. Also, for the downward case, it appears that you have taken x_naught = 0, but, for the downward case, x_naught is actually the maximum height of the bullet. 


#4
Aug806, 01:13 AM

P: 20

Vertical fall with quadratic air drag.
I tried flipping the signs but I still end up with a similar answer to my previous one, I admit I overlooked that minus, but still, this doesn't really help me reach my proof. Where do A and B come from? Perhaps a better question is, can someone tell me what A and B are so I know how to work to the proof. I've been stuck on this for a few days and it's starting to really annoy me.
I need help, Argh. Help me. Levi 


#5
Aug806, 03:59 AM

P: 20

Hey guys,
It's ok now, I finally sat down, cleared my mind of angst and frustration and carefully worked it out. It looks good now. Thank you both berkeman and George Jones. Levi 


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