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.