Question

Can somebody help me understand this equation? The graph of y(x) would be very insightful.

Answer 1

\[y(x)=(\frac{ gm }{ bv _{0} }\sec \theta+\tan \theta)x+\frac{ gm ^{2} }{ b ^{2} }\ln (\frac{ m }{ b }v _{0} \cos \theta -x)\]

Answer 2

www.wolframalpha.com

Answer 3

looks physic-y ... gravity, mass, otherstuff?

Answer 4

b0 beginning velocity ... not sure what a b would be

Answer 5

vo that is ... not sure what a b is

Answer 6

Yes. It what happened when I tried to solve the equations of motion for a projectile sent from the gound at an angle theta and with initial velocity vo. The gravitational field g is constant and there is air resistance (b) proportional to the velocity.

Answer 7

\[m\frac{ d ^{2}x }{ dt ^{2} }=-b \frac{ dx }{ dt }\]

Answer 8

\[m \frac{ d ^{2}y }{ dt ^{2} }=-mg-b \frac{ dy }{ dt }\]

Answer 9

hmm, i see that as x' = kx

Answer 10

well, x'' = kx' but thats just fat fingers

Answer 11

Well that's nice that the y one is just the nonhomogenous version of the x one.

Answer 12

ive got an exam today so my focus on other stuff isnt up to par .... good luck

Answer 13

So @ivancsc1996 I'm not sure what you want us to help you with. I gave you a link to wolram alpha so you can try graphing it there, but past that I don't think you've asked anything else?

Answer 14

I wanted you guys to plot the graph for me. I already tried wolfram but it doesn't seem to be able to plot y=ax+b ln (c-x)

Answer 15

The initial conditions are x(0)=0, y(0)=0, x'(0)=vocos theta y'(0)=vosen theta

Answer 16

http://www.wolframalpha.com/input/?i=plot+y%28x%29%3D8x%2B1ln+%285-x%29%2C+w%28x%29%3D4x%2B1ln+%285-x%29%2Cy%28x%29%3D8x%2B4ln+%285-x%29%2Cy%28x%29%3D8x%2B1ln+%2810-x%29

Answer 17

Thanks. But I must'v solved the equations badly. It doesn't look like anything