Question

A gun is fired straight up. assuming that the air drag on the bullet caries quadratically with speed, show that the speed varies with the height according to the equations:
v^2= A*e^(2kx) -g/k (upward motoin)
and v^2= g/k - B*e^(2kx) (downward motion)

A and B are constants of integraion, and k=C2/m, where C2 is the drag constant, and m is the mass of the bullet.

Answer 1

btw, I will give GOLD!

This is what I have done so far:
so I can find V(naught) as a function of X (position), there are my assumtions;
Up is positive, V(final)=0, and Xinitial=0.
I will call C2(drag constant) : c for simplicity.
ma=mg+cv^2
mdv/dt= mg+cv^2
mvdv/dx= mg+cv^2 (where dv/dt= (dv/dx)*(dx/dt))
and now:
mvdv/dx=mg(1+cv^2/mg)
vdv/dx=g(1+cv^2/mg)
(v/(1+cv^2/mg)) dv= gdx
after simplifying I get:
x= m/2c ln(1+cv^2/mg)

Now I don't know how to find x as a function of v^2 like in the suggested equations in the question.

Answer 2

This is hard to read, but if your last equation is correct:
\[ x= m/2c . \ln(1+cv^2/mg) \]
then
\[ \ln(1 + cv^2/mg) = 2cx/m \]
\[ cv^2/mg + 1 = e^{2cx/m} \]
\[ v^2 = \frac{mg}{c} (e^{2cx/m} - 1) \]

Answer 3

yeh but it does not match the answer they give for upward motion. I did it again now and I got this:
-mg/2c ln(1-cv^2/mg)=x
and when I solve for v^2
\[-2cx/mg= \ln(1-cv^2/mg)\]
\[-1+e^-2cx/mg =cv^2/mg\]
\[-mg/c+(mg/c)*e^{-2c/mg} = v^2\]
now, as told in the question k=c/m, then:
\[Ae^{-2k}-g/k = v^2\]
which is the same expression as the question where A=g/k
I think I got how to do the rest of the question, Thanks James