Projectile Motion in 2d with Drag

[nathanthegreat]

Homework Statement

So, I'm building a "Monkey Gun" and I want to be able to find the instantaneous velocity of my dart at a certain time. I shoot the gun at some angle, θ. The problem is, if I divide the velocity into x and y components, each depends on and impacts the other (the force from the direction will change the x velocity, changing the total velocity, changing the y velocity, etc.). I did a quick calculation assuming the dart remained at 60 m/s and I found a change in velocity of about 25 m/s, which is quite substantial. I know it won't actually be this much because the dart doesn't remain at 60 m/s, but it will still impact the dart.

I know I can mainly just ignore air resistance because the target is a decent size, but this is something I want to know for future experiments.

Homework Equations

$$F=ma$$ $$F_d =.5 \rho C_D A v^2$$

The Attempt at a Solution

I derived an equation for the instantaneous velocity for a freely falling object and got $$v(t)=√((mg-(mg-kv_0^2)e^((-2k)/m(∆t)))/k)$$. Now how can I find something like this in two dimensions? I know the y component will need to divided into a function for when the projectile is going up and a different one when it is coming down.

-Nathan

 

[haruspex]

You are quite right that if you take drag as proportional to the square of the speed then the horizontal drag is affected by the vertical speed, and v.v. There is no analytic solution; you'll need to use numerical methods. There are many references on the net. Some pretend drag is linear in order to get an analytic solution.

 

[rude man]

Why is it wrong to write
mx'' = - k(x')^2
my'' = -k(y')^2 - mg

these equations are uncoupled and it looks to me like they are solvable by
u = dx/dt
v = dy/dt
then separation of variables to get u and v, then solve the 1st order equations by separation again or some other way? Not that I've done it ...

Just asking, not asserting ...

 

[voko]

The force due to the drag is $-k v \vec{v}$.

 

[haruspex]

Why is it wrong to write
mx'' = - k(x')^2
my'' = -k(y')^2 - mg

The total drag is k(x'²+y'²). The component in the x direction is k(x'²+y'²) * x'/√(x'²+y'²) = kx'√(x'²+y'²).