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.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

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.

- JamesJ

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.