2D equation for projectile with linear drag force

[vsd50]

Homework Statement

Given the equations

a) find the solution to the problem (1) in vectorial form, by first writing equation (1) in component form and then solving the two parts separately. These can then be combined to obtain the vector form of the solution.

b) solve the results of the previous part to find x(t) and y(t).

c) taking m=1 kg, g=10 m/s^2 , plot the solutions y verus x for k=0, and k=1 on the same graph for 0<=t<=10000s (Use Maple)

Homework Equations

(1) dv/dt = -g(j-hat) - kv (where v is a vector)
(2) v= vx(i-hat) + vy (j-hat) = (dx/dt) (i-hat) + (dy/dt) (j-hat) (vy : velocity in the y, vx: velocity in the x)
(3) v(t=0) = v(o)= v1(i-hat)+v2(j-hat)

The Attempt at a Solution

so for part a I realized I could use equation 1 and 2 to form something like this
(d/dt)<dx/dt,dy/dt> = <0,-g> -k<dx/dt,dy/dt>
and then for the velocty in the x direction i get
d^2x/dt^2 = -k(dx/dt)
and for the velocity in the y I get
d^2y/dt^2 = -g -k(dy/dt)

Now my problem is that I am supposed to get this back into vectorial form would this be the way I do that?

x' = (-1/k)(x")
y' = (-1/k)(y"+g)

and then plug that back into this
v= vx(i-hat) + vy (j-hat)

so that I get

v=(-1/k)(x")(i-hat) + (-1/k)(y"+g)(j-hat)

also sorry for using a different notation for the derivatives at the end it just looked neater.

 

[Orodruin]

While you can write it like that, it is usually preferable to write an equation for the acceleration rather than for the velocity. You may also want to introduce the acceleration vector $\vec a = d\vec v/dt$.

Your results seem fine.

I also suggest that you look into using LaTeX to typeset your equations.

 

[vsd50]

ok so I did part a correctly so for part b I did something like this
$$ \ddot x = -k\dot x$$
$$ m^2+km = 0$$
$$ m_1=0$$ and $$m_2 = -k$$
then $$x=c_1 + c_2(e^{-kt})$$
and
$$\ddot y = -g -k\dot y$$
and seeing -g as $$-\ddot y$$
will yield a $$(2m^2+km)=0$$
where $$m_1=0$$ and $$m_2=\frac {-k}{2}$$
and then yields $$y=c_1+c_2(e^{\frac {-kt}{2}})$$

so how would I use equation (3) to find the constants in both y(t) and x(t) because in part c I need to graph them and also my professor mentions m= 1kg but I don't see a value for m anywhere in x(t) and y(t) so does that mean I am wrong?

 

[vsd50]

and thank you for your quick reply before sorry for not that earlier

 

[Orodruin]

and seeing -g as $$-\ddot y$$

Not sure what you mean by this. It is certainly not true that $\ddot y = -g$. If you assume this your differential equation will not be satisfied any time $\dot y \neq 0$. Your solution in the $x$-direction looks fine.

 

[vsd50]

so should I solve for y assuming that g is an Undetermined Coefficient and use a superposition approach?

 

[Orodruin]

so should I solve for y assuming that g is an Undetermined Coefficient and use a superposition approach?

Sounds reasonable. Why don't you try it?

 

[vsd50]

ok so I will treat g as an undetermined coefficient and use a superposition approach and use the $$y=y_c+y_p$$

this will then be $$\ddot y + k\dot y = -g$$

then $$y_c$$ will be

$$m^2+km=0 $$

$$m_1=0$$ and $$m_2=-k$$

then $$y=c_1+c_2(e^{-kt})$$

$$y_p=At $$

then $$\dot y_p = A$$ and $$\ddot y_p = 0$$ which yields

$$0+kA=-g$$ then $$A=\frac {-g}{k}$$

so since $$y=y_c+y_p$$

$$y=c_1+c_2(e^{-kt})-\frac {-gt}{k}$$

now that I have found x(t) and y(t) how do I find the constants using the equation (3).
Well i guess what I am asking is what do I do with x(t) and y(t) seeing as I am asked in part c to graph it in a y versus x graph and I am given values of k (k=0 and k=1), m=1kg and g=10m/s^2. Is this a parametric equation?

 

[vsd50]

Thanks for all of the help yesterday but I am still struggling to find values for all of the constants in x(t) and y(t) the only thing i get when t=0 is
$$v_1=c_1+c_2$$ and $$v_2=c_1+c_2$$

 

[Orodruin]

Well, clearly you need enough initial conditions in order to solve for the constants. Since they are second order differential equations, you will need initial conditions for both position and velocity. Note that the constant terms (i.e., the position at time $t = 0$) disappear for the initial condition on the velocities.