Projectile motion with air resistance proportional to velocity

[dykkms]

Homework Statement

The projectile with initial velocity $$v_0$$ is moving in environment with air resistance, which is linear to projectile's velocity. The task is to prove, that horizontal distance is maximal, if the elevation angle ($$\alpha$$) and the angle of trajectory tangent in place of impact are complementary.

Homework Equations

$$F_x = - k.v_x$$
$$F_y = - m.g - k.v_y$$
$$v_x = v.\cos(\varphi)$$
$$v_y = v.\sin(\varphi)$$
$$v_{0x} = v_0.\cos(\alpha)$$
$$v_{0y} = v_0.\sin(\alpha)$$

The Attempt at a Solution

equations in $$x$$ direction:
$$a_x = \frac{F_x}{m} = - \frac{k}{m} v_x$$
$$a_x = \frac{\mathrm{d}v_x}{\mathrm{d}t} = - \frac{k}{m} v_x$$
$$\frac{\mathrm{d}v_x}{v_x} = - \frac{k}{m}\mathrm{d}t \Rightarrow v_x = v_{0x}.e^{-\frac{k}{m}t}$$
after integration:
$$x = -\frac{m}{k} v_{0x} . e^{-\frac{k}{m}t} + \frac{m}{k} v_{0x}$$

equations in $$y$$ direction:
$$a_y = \frac{F_y}{m} = - g - \frac{k}{m} v_y$$
$$a_y = \frac{\mathrm{d}v_y}{\mathrm{d}t} = - g - \frac{k}{m} v_y$$
$$\frac{\mathrm{d}v_y}{- g - \frac{k}{m} v_y} = \mathrm{d}t$$
after integrations and mathematical expressing:
$$v_y = \frac{m}{k}\left(g + \frac{k}{m}v_{0y}\right)e^{-\frac{k}{m}t} - g\frac{m}{k}$$
$$y = - \frac{m^2}{k^2}\left(g+ \frac{k}{m}v_{0y}\right).\left(e^{-\frac{k}{m}t} - 1\right) - g\frac{m}{k}t$$

And here is my problem:
I would like to express from $$y$$ equation the time of the impact $$t_i$$. I know about this moment, that $$y = 0$$. If I am able to write explicit function for $$t_i(\alpha)$$, then I would pass it to $$x$$ function. Afterwards I would differentiate this $$x$$ function according to $$\alpha$$. Then I would find the ideal $$\alpha$$, which I would use to find $$t_i$$ which I would finally use to find impact angle.
Unfortunatelly, I can't see the way to find explicit expression of $$t_i$$ from $$y = 0$$ function.

 

[gabbagabbahey]

Basically, you want to maximize $x(t_i)-x(0)$ subject to the constraint $y(t_i)=0$...that seems like a prime candidate for the mathod of Lagrange multipliers to me

 

[dykkms]

Basically, you want to maximize $x(t_i)-x(0)$ subject to the constraint $y(t_i)=0$...that seems like a prime candidate for the mathod of Lagrange multipliers to me

Thank you for your answer!
Basically, you say (and I think you are right...), there are two other possibilities:

• from function $$y = 0$$ not to express $$t_i(\alpha)$$, but $$\alpha(t_i)$$ and to use this in function $$x(t_i)$$, which I should maximize
• to use method of Lagrange multipliers

First mentioned method brings me to situation, that I am not able to express $$t_i$$, which would maximize $$x(t_i)$$ function, anyway.
Second, I was thinking about method of Lagrange multipliers before, but I tried to avoid it, because I am trying to help with this homework to my friend, who is first-year student at university (industrial management) and he doesn't know the method yet... But I think, I will have to.

 

[gabbagabbahey]

Hmmm... if this is supposed to be a 1st year level problem, there may be a simpler method...perhaps energy conservation would make things easier.

 

[paranormal]

First of all, great job on setting up the equations for projectile motion with air resistance! To find the time of impact, we can use the fact that at the impact point, the projectile's position in the y-direction is 0. This means that we can set y = 0 in the y-equation and solve for t.

y = - \frac{m^2}{k^2}\left(g+ \frac{k}{m}v_{0y}\right).\left(e^{-\frac{k}{m}t} - 1\right) - g\frac{m}{k}t
0 = - \frac{m^2}{k^2}\left(g+ \frac{k}{m}v_{0y}\right).\left(e^{-\frac{k}{m}t} - 1\right) - g\frac{m}{k}t
\frac{m^2}{k^2}\left(g+ \frac{k}{m}v_{0y}\right).\left(e^{-\frac{k}{m}t} - 1\right) = - g\frac{m}{k}t
\left(e^{-\frac{k}{m}t} - 1\right) = - \frac{gk}{m(g + \frac{k}{m}v_{0y})}t

We can then use the natural logarithm to solve for t:

\ln\left(e^{-\frac{k}{m}t} - 1\right) = \ln\left(- \frac{gk}{m(g + \frac{k}{m}v_{0y})}t\right)
-\frac{k}{m}t = \ln\left(- \frac{gk}{m(g + \frac{k}{m}v_{0y})}\right) + \ln t
t = \frac{m}{k}\left(\ln\left(- \frac{gk}{m(g + \frac{k}{m}v_{0y})}\right) + \ln t\right)

Since we want to find the time of impact as a function of the initial velocity and angle, we can substitute in the expressions for v_{0x} and v_{0y} in terms of v_0 and \alpha:

t = \frac{m}{k}\left(\ln\left(- \