How can I quickly destroy the enemy target and avoid being discovered?

[MarkFL]

You are a gunner stationed on a distant outpost located on a desolate planet with no atmosphere. When you signed up for duty, you envisioned traveling the galaxy and seeing many wondrous things. Your recruiter painted quite an exciting picture, regaling you with many fascinating stories of his time in deep space! You think of the irony as you stare at the walls of your sparsely decorated dorm room day after day, each of which lasts roughly 3 Earth weeks on this stupid planet. (Dull)

But, this planet is rich in minerals that mankind wishes to mine, and so here you are. However, another race of beings also wants to mine this planet, and so naturally we are at war, as the universe is simply too small to share. (Punch)

Your gun is underground, but can maneuver such that its muzzle is at ground level for any launch angle $0<\theta<\dfrac{\pi}{2}$, and the terrain is flat all around your location. Your gun has a muzzle velocity of $v_0$ and the acceleration due to gravity is $g$.

Suddenly your boredom is shattered as an enemy craft has just unwittingly landed within the maximum range of your gun. If $r_{\max}$ is the maximum range of your gun, then let the range of the enemy be $k\cdot r_{\max}$, where $0<k<1$. It has been determined that the EM shielding of the enemy craft is capable of withstanding being hit by one of your projectiles, however, if hit by two projectiles simultaneously, the shield will collapse and the enemy will be destroyed. If you fail, your location will have been given away, and the enemy will begin firing on your location. If you take too long, the enemy will also inevitably discover your location and begin firing. Thus, you need to quickly and correctly determine:

a) The two launch angles for the projectiles.

b) How long after the first projectile is fired should the second be fired?

 

[I like Serena]

Find the right angle to offer flowers to your enemy.
Any enemy should be appeased by love (Inlove) and flowers (flower).
We should all be able to live together in peace and harmony. (Hug)

 

[Deveno]

Well obviously the "right" angle is $\frac{\pi}{2}$...

Looking at my terribly messy calculations, I think I have:

$\theta_2 = \frac{1}{2}\text{arcsin}\left(\dfrac{gk r_{\text{max}}}{v_0^2}\right)$

and $\theta_1 = \frac{\pi}{2} - \theta_2$

with:

$\Delta t = t_1 - t_2 = \dfrac{2v_0}{g}(\sin\theta_1 - \cos\theta_1)$

(the first shot has to be fired at an angle > $\frac{\pi}{4}$ to have a longer "hang time" which only depends on the vertical component)

but I am too tired at this point to check my work.

 

[MarkFL]

Hello, Deveno!

Your answers are correct and you have destroyed the enemy!

My results were in a different form, and are given here:

Spoiler

Along the horizontal component of motion, there are no forces acting on the projectile, so we may state:

\(\displaystyle \frac{dv_x}{dt}=0\) where \(\displaystyle v_{x_0}=v_0\cos(\theta)\)

Integrating with respect to $t$, we find:

\(\displaystyle v_x(t)=v_0\cos(\theta)\)

Integrating again, where the origin of our $xy$-axes is at the muzzle, we find:

\(\displaystyle x(t)=v_0\cos(\theta)t\)

Along the vertical component of motion the force of gravity is acting, in a downward direction, so we have:

\(\displaystyle \frac{dv_y}{dt}=-g\) where \(\displaystyle v_{y_0}=v_0\sin(\theta)\)

Integrating with respect to $t$, we find:

\(\displaystyle v_y(t)=-gt+v_0\sin(\theta)\)

Integrating again, we get:

\(\displaystyle y(t)=-\frac{g}{2}t^2+v_0\sin(\theta)t\)

Eliminating the parameter $t$, we obtain:

\(\displaystyle y(x)=\tan(\theta)x-\frac{g}{2v_0^2\cos^2(\theta)}x^2\)

Equating $y$ to zero and taking the non-zero root, we get the range $r$ of the projectile as:

\(\displaystyle r=\frac{v_0^2\sin(2\theta)}{g}\)

We see now that the range is a maximum for \(\displaystyle \theta=\frac{\pi}{4}\) and is given by:

\(\displaystyle r_{\max}=\frac{v_0^2}{g}\)

And so the range of the enemy is:

\(\displaystyle r=k\frac{v_0^2}{g}=\frac{v_0^2\sin(2\theta)}{g}\)

This then implies:

\(\displaystyle \sin(2\theta)=\sin(\pi-2\theta)=k\)

Thus, the larger launch angle (which will be used for the first projectile as it will be in motion longer) $\theta_1$ is found from:

\(\displaystyle \pi-2\theta_1=\sin^{-1}(k)\)

\(\displaystyle \theta_1=\frac{\pi}{2}-\frac{1}{2}\sin^{-1}(k)\)

And of course the smaller launch angle used for the second projectile is:

\(\displaystyle \theta_2=\frac{1}{2}\sin^{-1}(k)\)

Now, using the equation:

\(\displaystyle r=v_0\cos(\theta)t\implies t=\frac{r}{v_0\cos(\theta)}=\frac{k\frac{v_0^2}{g}}{v_0\cos(\theta)}=\frac{kv_0}{g}\sec(\theta)\)

And so we find, using the fact that the two launch angles are complementary:

\(\displaystyle \Delta t=t_1-t_2=\frac{kv_0}{g}\left(\csc\left(\frac{1}{2}\sin^{-1}(k) \right)-\sec\left(\frac{1}{2}\sin^{-1}(k) \right) \right)\)

Using half-angle identities for sine and cosine, and simplifying a bit, we obtain:

\(\displaystyle \Delta t=\frac{\sqrt{2}v_0}{g}\left(\sqrt{1+\sqrt{1-k^2}}-\sqrt{1-\sqrt{1-k^2}} \right)\)

 

[topsquark]

Hello, Deveno!

Your answers are correct and you have destroyed the enemy!

Spoiler

\(\displaystyle \Delta t=\frac{\sqrt{2}v_0}{g}\left(\sqrt{1+\sqrt{1-k^2}}-\sqrt{1-\sqrt{1-k^2}} \right)\)

Well it's good to know that I got the answer correct, but I was too lazy to work out the half-angle formulas.

-Dan