# Rapid-fire projectile motion such that all particles land at the same time

A short-barrelled machine gun stands on horizontal ground. The gun fires bullets, from ground level, at speed $$u$$ continuously for a period of one second, but does not fire outside of this interval.

During this time period, the angle of elevation of the barrel, $$\theta(t)$$, decreases from $$\frac{\pi}{4}$$ to $$\frac{\pi}{6}$$.

How can I find the function for $$\theta(t)$$ that will ensure all the bullets land at the same time?

(Assumptions: no air resistance, no recoil of the gun between shots, ground is level, bullets have no size)

This is a somewhat made-up problem. I'm not too sure about how to start, or even if it is possible for them to land at the same time. It's clear that $$\theta(0) = \frac{\pi}{4}$$ and $$\theta(1) = \frac{\pi}{6}$$ and the result $$t = \frac{2u \sin \theta}{g}$$ will probably be useful, but other than that I'm stuck.

Any help much appreciated!

Edit: since asking the question initially I think I've been able to make some progress.

Take the start of the first bullet being fired as $$t = 0$$. Then the time when one of the other bullets lands satisfies

(time delay to fire previous bullets) + (time of flight of this bullet) = (time of flight of first bullet).

$$t + \frac{2u \sin \theta}{g} = \frac{2u \sin \frac{\pi}{4}}{g}$$

Differentiating with respect to $$t$$,

$$1 + \frac{2u}{g} \cos \theta \times \frac{d\theta}{dt} = 0$$

But WolframAlpha gives the solution (with $$\theta(1) = \frac{\pi}{6}$$ for the last bullet as the boundary condition) as

$$\theta(t) = \sin^{-1} (\frac{u - gt + g}{2u})$$

which does not even pass through $$(t = 0, \theta = \frac{\pi}{4})$$.

What went wrong?

• Re, "...or even if it is possible..." should be obvious that it is possible. The time for a bullet to return to the ground will depend only on the vertical component of its initial velocity, and you reduce that vertical component when you reduce the angle of the gun. Jul 22, 2020 at 13:15
• The gun will fire a small, whole number of bullets during that one second. You can calculate the time when the first bullet hits the Earth. Then for each next bullet/each later starting time $t$, work backward to find the necessary angle, $\theta{}(t)$, to make the bullet land simultaneously with the others. Now you have a table of values of $\theta{}(t)$ for some ten or twenty $t$. You should be able to fit a continuous function to those values. Worst case: you can always fit a polynomial whose order is the same as the number of rows in your table. Jul 22, 2020 at 13:20
• This is a theoretical gun and so I imagine it firing a huge number (approaching infinity) of bullets with no time difference between them. This gives a continuous stream of bullets. Jul 22, 2020 at 13:22
• Since asking the question I have made some progress, being able to form a differential equation for $\alpha(t)$ but some graphing of the results shows it doesn't work. Would it help if I post what I have so far? Jul 22, 2020 at 13:23
• BTW Time On Target en.wikipedia.org/wiki/Time_On_Target is a real thing for both multiple guns and multiple shots from one gun. Nice graphic on Wikipedia Jul 22, 2020 at 22:10

I was able to find the issue and apparently (according to meta) its OK to post an answer to my own question.

The working in the question leading to the solution of

$$\alpha(t) = \sin^{-1} \left (\frac{1}{2} + \frac{g}{2u}(1-t) \right )$$

is correct, but it's only possible for the first and last bullet to land at the same time for a particular velocity. This was found by using the boundary conditions to verify the first equation, $$1 + \frac{2u \sin \frac{\pi}{6}}{g} = \frac{2u \sin \frac{\pi}{4}}{g}$$

which can be solved for $$u$$ to get

$$u = (1 + \sqrt{2})g.$$

So the only actual equation that makes it work, and does so only in the case of this initial speed, is

$$\alpha(t) = \sin^{-1} \left (\frac{1}{2} + \frac{1-t}{2(1+\sqrt{2})} \right ).$$

I got this solution ?

with:

$$y=v_0\,\sin(\varphi)\,t-\frac{g\,t^2}{2}\tag 1$$

$$\Rightarrow$$

$$y=0\quad,t_0(\varphi=\varphi_0)=2\,{\frac {v_{{0}}\sin \left( \varphi _{{0}} \right) }{g}}$$

for next bullet: $$t\mapsto t_0-dt$$ and $$\varphi\mapsto \varphi(t)$$ in equation (1)

$$y_t=v_{{0}}\sin \left( \varphi \left( t \right) \right) \left( 2\,{ \frac {v_{{0}}\sin \left( \varphi _{{0}} \right) }{g}}-{ dt} \right) -\frac 1 2\,g \left( 2\,{\frac {v_{{0}}\sin \left( \varphi _{{0}} \right) }{g}}-{ dt} \right) ^{2}\tag 2$$

we solve $$y_t=0$$ and get $$dt=(dt_1\quad,dt_2)$$

$$dt_1=t_0\quad dt_2=dt$$ $$dt-2\,{\frac {v_{{0}} \left( -\sin \left( \varphi \left( t \right) \right) +\sin \left( \varphi _{{0}} \right) \right) }{g}} =0\tag 3$$

take the time derivative of equation (3):

$$1-2\,{\frac {v_{{0}}\cos \left( \varphi \left( t \right) \right) { \frac {d}{dt}}\varphi \left( t \right) }{g}} =0$$

solve this equation for $$\dot\varphi\quad \Rightarrow$$

$${\frac {d}{dt}}\varphi \left( t \right) -\frac 1 2\,{\frac {g}{v_{{0}}\cos \left( \varphi \left( t \right) \right) }} =0$$

thus the solution

$$\varphi(t)=\arcsin \left( \frac 1 2\,{\frac {g \left( t+{C} \right) }{v_{{0}}}} \right) \quad ,\text{with}\quad C=t_0$$

$$\varphi(t)=\arcsin \left( \frac 1 2\,{\frac {gt+2\,v_{{0}}\sin \left( \varphi _{{0}} \right) }{v_{{0}}}} \right) \tag 4$$

with $$\varphi(t=1)=\varphi_e$$ you can obtain the start velocity $$v_0$$

$$v_0=-\frac 1 2\,{\frac {g}{\sin \left( \varphi _{{0}} \right) -\sin \left( \varphi _{{e}} \right) }}> 0 ,\quad \Rightarrow\quad \varphi_e > \varphi_0$$

• this seems to work, thanks for showing the more general technique :) Jul 22, 2020 at 20:42

I dont know why you took the derivative (plus I think you might have a typo in wolfram alpha), but after you get $$t + \frac{2u \sin \theta}{g} = \frac{2u \sin \frac{\pi}{4}}{g}$$, you can get $$\theta (t)$$ straight from there as:

$$\theta (t)=\sin^{-1}(\sqrt{2}/2-gt/2u)$$

this is a continuous function and passes through $$\pi /6$$, when $$t=(\sqrt {2}-1)u/g$$

• Idk...I thought the answer would be fairly complicated so would need a differential equation (so i differentiated to make one). Surprised to see this wasn't necessary! Jul 22, 2020 at 20:36
• you still should get the right answer though, that is why I mentioned there might be a typo somewhere when you got the Wolfram solution
– user65081
Jul 22, 2020 at 20:37
• yeah i just graphed yours on desmos and its the same :) I don't think there's an error in wolfram, I got out the same thing as you, I just switched the letters g, u and $\theta$ for a, b and $x$ Jul 22, 2020 at 20:39
• I actually get $u=g/(\sqrt{2}-1)$, which is actually the same you got
– user65081
Jul 22, 2020 at 20:47
• you edited after I posted, anyway its resolved. thanks for your solution Jul 22, 2020 at 20:54