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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.

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# Vertical fall with quadratic air drag.

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