Projectile motion with air resistance

[matineesuxxx]

Homework Statement

Model the trajectory of a projectile with an initial velocity $v_0$ that is subject to the force of drag.

Homework Equations

$F_{\text{d}} = \frac{1}{2}C_{\text{d}}\rho\text{A}(v(t))^2$

$v_x(t) = v_{0x} + \int_{t_0}^t a_x(t')\mathrm{d}t'$

The Attempt at a Solution

I really have no clue how to tackle this problem. I'm sort of leaning towards the idea of a differential equation solution, however I only have a small amount of DE's under my belt from calc 2.

I have done a fair amount of searching online, however every page that I've seen about this problem assumes drag force is linearly dependent on speed, or even assuming drag is constant. Would I gain enough insight solving the problem in which drag force is linear to make headway on the real life scenario?

 

[mfb]

If you replace F in your first equation with one of Newton's laws, you get a differential equation with the velocity and its time-derivative.

 

[rcgldr]

You need to express acceleration as an equation. mfb gave you a hint for how to do this.

Then note that acceleration = dv/dt, but in this case, acceleration will be a function of velocity:

dv/dt = a(v)

to deal with this rearrange it to:

dv/a(v) = dt

then integrate both sides to get velocity versus time (don't forget to include a constant of integration after integrating). You'll need to repeat this process again and integrate to get position versus time.

 

[matineesuxxx]

Thanks a lot, both of you. It's been a while since I've done DE's, so I didn't see what now seems so obvious!

So I get,

$F_{\text{d}} = -\frac{1}{2}C_{\text{d}}\rho \text{A}v^2 = m\frac{\text{d}v}{\text{d}t}$

which then leads me to,

$$ v(t) = \frac{2m}{K_1 + C_{\text{d}}\rho \text{A}t}$$,

where $K_1 = C_{\text{d}}\rho \text{A}mK_0$ and $K_0$ is just the constant of integration.

 

[matineesuxxx]

You'll need to repeat this process again and integrate to get position versus time.

when I use $$ x(t) = x_0 + \int_{t_0}^{t_1} \frac{2mv_{0x}}{C_d \rho A v_{0x}t + 2m}dt $$

to get my displacement vs time graph I get $$x(t) = \frac{2m}{C_d \rho A}\ln\left(\frac{2m+C_d\rho A v_{0x}t}{2m}\right) $$

but this reduces to 0, not $x(t) = v_{0x}t$ if there is no drag. So is this correct, or have I made a mistake?

[edit] Sorry, I neglected to take a limit as c, p, or A went to zero

 

[matineesuxxx]

I'm thinking velocity should be:

$$ v(t) = \frac{1}{\frac{1}{v_0} - \frac{t}{C_{\text{d}}\rho \text{A}}}$$

Note that at t = 0, v = v0,and at t = ∞, v = 0.

Here's what I did:

let $\mu = C_d \rho A$ so then $-\frac{1}{2}\mu v^2 = m\frac{dv}{dt} \implies v\mu t = 2m - vK$.

so $v = \frac{2m}{\mu t + K}$, where $v(0) = v_{0x} \implies K = \frac{2m}{v_{0x}}$ so then I end up with $$ v = \frac{2mv_{0x}}{2m+\mu v_{0x}t} = \frac{2mv_{0x}}{2m + C_d \rho A v_{0x}t}$$

which still reduces to the correct equations when taking limits, and we don't have a nasty asymptote.

 

[rcgldr]

Sorry for the previous post, I wasn't paying attention to the fact that the force opposes velocity and should be negative: m dv/dt = -1/2 u v^2 . What you have looks correct.

If the drag is zero, then the integral using CdpA to produce x(t) is undefined as you end up with something in the form of (1/0) ln(1) => 0/0. So if CdpA == 0, the integral becomes:

$$ x(t) = x_0 + \int_{t_0}^{t_1} \frac{2mv_0}{0 + 2m}dt $$
$$ x(t) = x_0 + \int_{t_0}^{t_1} v_0dt $$

 

[matineesuxxx]

Sorry for the previous post, I wasn't paying attention to the fact that the force opposes velocity and should be negative: m dv/dt = -1/2 u v^2 . What you have looks correct.

If the drag is zero, then the integral using CdpA to produce x(t) is undefined as you end up with something in the form of (1/0) ln(...). So if CdpA == 0, the integral becomes:

$$ x(t) = x_0 + \int_{t_0}^{t_1} \frac{2mv_0}{0 + 2m}dt $$
$$ x(t) = x_0 + \int_{t_0}^{t_1} v_0dt $$

No worries, I actually forgot to add the negative sign in the first occurrence of the force equation, so i made the same mistake :). And in regards to the (1/0)ln(...), it ends up as 0/0 so just use L'Hopital's Rule and you still end up with x(t) = (v_0x)t, same goes for m -> infinity

 

[matineesuxxx]

Thanks for all the help! Here's a neat interactive graph I did up for it:

https://www.desmos.com/calculator/28tlkhbgn9

 

[rcgldr]

Thanks for all the help! Here's a neat interactive graph I did up for it:

So you enhanced the equations to a 2d situation? I'm not sure I understand the graphs. I assume the graph shows the X Y plane with gravity exerting a force in the -Y direction and drag always opposing velocity (both x and y components), however the peak height reached in the graphs for the object remains the same regardless of drag. It appears the calculator adjusts the initial angle of velocity to do this, but it would seem that if drag is sufficiently large, the peak would be less than the zero drag case, even if the initial velocity was vertical.

 

[matineesuxxx]

The maximum height reached would not change given any air drag, so long as the drag force is purely horizontal, which is one condition of my problem, although I neglected to mention that. And you the graphs are a trajectory, horizontal vs vertical distance.

 

[rcgldr]

The maximum height reached would not change given any air drag, so long as the drag for is purely horizontal.

Ok, so an imaginary case where drag is only horizontal. I was expecting something like this:

http://demonstrations.wolfram.com/ProjectileWithAirDrag

It might be interesting to plot a postion versus time graph (straight line for zero drag non-zero velocity).

 

[matineesuxxx]

Ok, so an imaginary case where drag is only horizontal. I was expecting something like this:

http://demonstrations.wolfram.com/ProjectileWithAirDrag

It might be interesting to plot a postion versus time graph (straight line for zero drag non-zero velocity).

Yes. I wanted to originally have drag act as it really would, but I have no clue how to solve a differential equation of the form $q(x) = \frac{dy}{dx} + p(x)y^2$.

 

[rcgldr]

That wolfram article is using an imaginary drag factor that varies with velocity as opposed to velocity squared.

I though about parametric approach, but apparently the equations involve v and vx and vy:

$$ m\ \frac{dv_x}{dt} = - \ c\ v \ {v_x} $$
$$ m\ \frac{dv_y}{dt} = -m\ g \ - \ c\ v \ {v_y} $$

based on this pdf article that uses numerical integration to solve the problem:

topic01.pdf

If the object is falling vertically, then there's no x component, and the equations can be solved:

wiki free fall with air resistance .htm

 

[mfb]

Yes. I wanted to originally have drag act as it really would, but I have no clue how to solve a differential equation of the form $q(x) = \frac{dy}{dx} + p(x)y^2$.

Do it numerically.