Question

Can somebody help me understand this equation? The graph of y(x) would be very insightful. I arrived to it by trying to analyze the motion of a particle under the force of a constant gravitational field and a resistance proportional to its velocity. ie: parabolic motion with air resistance

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

I've attached a plot from Matlab. The red plot is plotted at θ=1°. The green is plotted at θ=10°, and the blue plot is done at θ=25°. I did not enter values for g, m, b, or v0.

Answer 3

If you are 'just playing' You would be better off with resistance proportional to v^2

For motion through air bodies have air resistance =
F = Cd * A * rho * v^2/2

Cd = coefficient of drag (typically 0.5 - 1 depending on shape of oblect)
A = characteristic area - typically frontal area of body
rho = density of air
v = velocity

http://en.wikipedia.org/wiki/Drag_equation

Answer 4

Wow that doesn't look like any possible motion of a projectile. I solved the equations \[m\frac{ d ^{2}x }{ dt ^{2} }=-b \frac{ dx }{ dt }\]\[m \frac{ d ^{2}y }{ dt ^{2} }=-mg-b \frac{ dy }{ dt }\] Maybe I did something wrong? To solve the "x" equation I just used changed the equation to the characteristic equation and solved it like a linear homegeneous second order ODE. For the second equation I head on integrated with respect to t and the solved the linear first order ODE making use of integration factors.

Can you please explain me why it would be better to use the \[F _{D } \alpha v ^{2}\]insted of\[F _{D} \alpha v\]

Answer 5

@ivancsc1996
I explained it in my earlier posting

The resisting force on a projectile in horizontal direction is due to air resistance.

the drag force is proportional to square of velocity.

The link gives you the equation - but it is well documented in fluid dynamics.

Answer 6

This is of course true in vertical component too, with the added constant vertical force of the weight (mg)

note that it is (dx/dt)^2, nor (d2x/dt^2)

I think my calculus is too rusty to attempt to help with the solution - but I am confident that v^2 is a better model of resistance force

Answer 7

Wikipedia has the solution to this (scroll down for the 'with air resistance' section)

http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

It works on F= kv for low velocities, but states that f= kv^2 is more complete

It also states that f=kv^2 does not have an analytical solution

The equations you are after are solved in that link