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Set the initial position of the superball be at the origin on the Cartesian plane. The set of projectile equations is $\begin{array}{rl} h(t) &= -\dfrac{1}{2}gt^2 + v\sin\theta t\\ d(t) &= v\cos\theta t \end{array}$ where $\theta$ denotes the launch angle, $g$ the gravity and $v$ the launch velocity. Combining these equations altogether, we have $h(d) = -\dfrac{1}{2} \dfrac{gd^2}{v^2\cos^2 \theta} + d\tan\theta$

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Bouncing Function

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For each Step $n$ of the staircase, the position of its midpoint is $(n - 1, n + 1)$ . At Step $3$ and $6$ , their respective positions are $(2,-2)$ and $(5,-5)$ . Since there is no change in the ball's path, we can assume that the ball travels in the similar manner for each bounce. In this case, we can use the above function to construct the piecewise function of the bounces.

For the first bounce, we have the graph of $h(d)$ bounded at $0 \leq d \leq 2$ . Substituting the first coordinates $(2,-2)$ , we have $1 + \tan\theta = \dfrac{g}{v^2 \cos^2 \theta}$ For the second bounce, we determine another part of the function by symmetry about $d = d_{1f} = 2$ , which is the endpoint of the first bounce. Since the reflection of $h(d)$ about the vertical is $h(2d_{1f} - d)$ , then the equation of the second bounce is $h^{*}(d) = h(4 - d) = -\dfrac{1}{2}\dfrac{g(4 - d)^2}{v^2\cos^2\theta} + (4 - d)\tan\theta$ where if $h^{*}(5) = -5$ , $\begin{array}{rl} -5 &= -\dfrac{1}{2}\dfrac{g}{v^2\cos^2\theta} - \tan\theta\\ -5 &= -\dfrac{1}{2}\left(1 + \tan\theta\right) - \tan\theta \end{array}$ So $\tan\theta = 3$ , where $0^{\circ} < \theta < 90^{\circ}$ . With some algebra and trigonometry, we discover that the equations of two first bounces are $h(d) = \left\{ \begin{array}{rl} -2d^2 + 3d & 0 \leq d \leq 2\\ -2(4 - d)^2 + 3(4 - d) & 2 < d \leq 5 \end{array} \right.$ where the endpoint of the second bounce is exactly at $(5,-5)$ . By the similar method, the equation of the third bounce is $h^{**}(d) = -2(d - 6)^2 + 3(d - 6)$ where it is bounded at $5 < d \leq 9$ . Since $h^{**}(9) = -9$ , this shows that the ball $\boxed{\text{lands on the midpoint of Step 10}}$ .

Note: The equation verifies that since the ball misses Step $2$ and instead lands on the midpoint of Step $3$ , there is no calculation conflict in missing any step.

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Bonus

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Since the reflection about the vertical affects the values of $d$ , the general equation of the bounce is $h\left(D_b\right) = -2\left(D_b\right)^2 + 3D_b$ where for each positive bounce integer $b$ ,

• the bounds are $\frac{b(b + 1) - 2}{2} \leq d \leq \frac{(b + 1)(b + 2) - 2}{2}$
• $D_b = d(-1)^{b - 1} + \sum\limits_{n=1}^b (-1)^n \left(n^2 + n - 2\right)$ is the recurrence relation

The numbered step pattern is that

For each bounce integer $b \geq 2$ , given that the initial position is at the midpoint of Step $1$ , the ball exactly lands on the midpoint of Step $\frac{b(b + 1)}{2}$ .

The proof, which involves mathematical induction , is left for the readers to work out.