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# Crazy Cannons

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

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- Problem
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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