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Weekly Challenge 33: Crazy Cannons

Stage: 5 Short Challenge Level: Challenge Level:2 Challenge Level:2

Put the first cannon at the origin $(0, 0)$ and the second cannon at the point $(D, 0)$.

Using a constant acceleration of $-g$ in the $y$-direction and $0$ in the $x$-direction it is a simple matter to write down the positions of each cannon ball at a time $t> T$ if we make use of the formula $s=ut+\frac{1}{2}at^2$. 
 
$$
\begin{eqnarray}
(x_1, y_1)&=&\left(100\cos(45^\circ)t, 100\sin(45^\circ)t-\frac{1}{2}gt^2\right)\cr
(x_2, y_2)&=&\left(D-100\cos(30^\circ)(t-T), 100\sin(30^\circ)(t-T)-\frac{1}{2}g(t-T)^2\right)
\end{eqnarray}
$$
As with all mechanics problems, the first part involves a careful setup of the equations. Once I have checked these carefully (... OK, that's done...) we can proceed with the algebra to resolve the equations.
 
Since I know that the two cannon balls strike each other the plan of attack is to equate the two $x$ and $y$ coordinates. I find that
$$
50 \sqrt{2}t = D-50\sqrt{3}(t-T)
$$
and
$$
50\sqrt{2}t-5t^2=50(t-T)-5(t-T)^2\;.
$$
After some rearrangement, the second of these equations gives me
$$
\begin{eqnarray}
\left(10(\sqrt{2}-1)-2T\right)t &=& -T^2-10T\cr
\Rightarrow t = \frac{T^2+10T}{2T-10(\sqrt{2}-1)}\;.
\end{eqnarray}
$$
Since for a collision to occur we must have $t> 0$, which implies that
$$
T> 5(\sqrt{2}-1)\;.
$$
Thus, there is a minimum value of $T$ (which might be greater than $5(\sqrt{2}-1)$; it is not less than this value). Now, for a collision to occur in the air the $y$ coordinate at the point of collision must be positive. The expression for the first cannon ball quickly gives us the inequality
$$ t< 10\sqrt{2}\;.$$  
 
This gives us a more complicated inequality for $T$ as
$$
 \frac{T^2+10T}{2T-10(\sqrt{2}-1)}< 10\sqrt{2}\;.
$$
 Rearranging we see that
$$T^2+10(1-2\sqrt{2})T+100(2-\sqrt{2})< 0\;.$$
Values of $T$ which satisfy this equation are those lying between the two roots
$$
T_{1, 2} = \frac{10(2\sqrt{2}-1)\pm\sqrt{(10(1-2\sqrt{2})^2-4(100(2-\sqrt{2}))}}{2}\;.
$$
Thus,
$$
10(\sqrt{2}-1) < T< 10\sqrt{2}\;.
$$
I used a spreadsheet to plot the values of $D$ against $T$. The range of permissible values is