00:01
In this problem, we want to derive the linear drag formula, which is x equal to v0 over k, 1 minus e to the negative kt, cosine alpha, and y equal to b0 over k, 1 minus e to the negative kt, sine alpha, plus g over kt 2 minus kt, minus kt, minus kt, minus kt, minus kt, minus kt, and minus e to the negative kt.
00:37
So we want to find this from the initial value problem, d squared r over dt squared equals negative gj minus k, dr, dt.
00:54
So that's our ode.
00:57
And then we have initial conditions, r of zero is zero, and drdt of zero.
01:05
Is v .0 the vector, which is going to be, so v .0 cosine alpha i plus v .0 sine alpha j.
01:19
Okay, so let's go ahead and just break up this into two problems.
01:26
So one for x and one for y.
01:27
So we'll have x double prime is equal to negative k x prime.
01:38
We can rewrite that.
01:40
As x double prime plus k x prime is equal to zero.
01:48
And so what are the initial conditions being? so we'll have x of zero is zero, and x prime of zero is going to be v.
02:00
Not cosine alpha.
02:03
And then we'll do the same for y.
02:05
So we'll have y double prime is equal to y double prime is equal to negative g minus k y y or we can write that as y double prime plus ky is equal to negative g.
02:21
And the initial condition, so we'll have y of zero is zero, and y prime of zero is going to be v not sine alpha.
02:29
So that's our other problem.
02:31
So notice that x has a homogeneous problem and y has a non -homogeneous problem.
02:36
So we'll start with x because we're going to do similar steps when we do y, because y will have to split up into the homogeneous in particular solutions.
02:44
Okay, so for x, right, we have, so we want to find the characteristics.
02:53
So we'll have lambda squared plus k lambda is equal to zero.
02:58
So when we solve for lambda, we'll get lambda is equal to zero and negative k.
03:03
Just by factoring on lambda, you get lambda times lambda plus k.
03:07
All right, so that means that x looks like, so x of t is going to look like some constant times e to the negative kt.
03:16
Plus some other constant times e to the 0, which is going to be 1.
03:21
So we'll just have that constant like that.
03:24
So then we know that x of 0 is equal to 0, which is going to be equal to c1 plus c2.
03:32
So that means that c1 equals negative c2, whatever that may be.
03:37
We'll find out a second.
03:38
So then x prime, so if we take the derivative of this, d, t, d, t, we would get negative k, c1, e to the negative k t plus zero.
03:53
So we know that x prime of zero is equal to v .0 cosine alpha.
04:01
And that's going to be, so if we plug in zero into this equation, we'll just get negative k, c1.
04:07
So that means that c1 is equal to negative v0 over k cosine alpha.
04:16
All right, so what is x? so x of t is going to be, so we have z1, or sorry, c1, so negative v0 over k cosine times e to the negative kt.
04:32
And then we have plus c2.
04:34
So c2 is negative c1, so we'll have plus v0 over k cosine, and we can rewrite this, so we'll factor out the, so we can factor out.
04:47
The v0 over k times cosine alpha and then tidy it up a bit so we have that in both terms so we'll have v not over k cosine alpha times so here's a positive one and this will be a negative e to the negative k t all right it looks pretty good that's what we had here x makes sense uh y's not as much fun but well i mean it's just fun if you like it but see if i can copy this whole thing there we go.
05:26
Here, oh, gosh.
05:34
Similarly to x, we're going to have the same characteristics.
05:37
So we're going to have the y homogeneous is going to be some constant e to the negative kt plus some other constant...