### Weekly Challenge 43: A Close Match

Can you massage the parameters of these curves to make them match as closely as possible?

### Weekly Challenge 44: Prime Counter

A weekly challenge concerning prime numbers.

### Weekly Challenge 28: the Right Volume

Can you rotate a curve to make a volume of 1?

# Weekly Challenge 33: Crazy Cannons

##### Stage: 5 Short Challenge Level:

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