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?
 A: 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
$$ \alpha(t)  = \sin^{-1} \left (\frac{1}{2} + \frac{1-t}{2(1+\sqrt{2})}  \right ). $$
A: 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 $v_0$
$$v_0=-\frac 1 2\,{\frac {g}{\sin \left( \varphi _{{0}} \right) -\sin \left( 
\varphi _{{e}} \right) }}> 0 ,\quad \Rightarrow\quad \varphi_e > \varphi_0
$$
A: 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 $t=(\sqrt {2}-1)u/g$
