How much interesting mathematics could there be in such a simple problem, you ask unwittingly.

At time t=0, we drop the ball from a height h=$h_i$, measured from the mean position of the table to the bottom of the ball.  We assume that the ball drops under uniform accelleration g; then the position of the ball is
    h=$h_i$-$\frac{1}{2}$g $t^2$
    v(t)=-g t.

The table is oscillating with frequency ω, phase φ, and amplitude A about h=0:
    H=A cos(ω t-${\varphi }_i$)
    V(t)=-ω A sin(ω t -${\varphi }_i$).

The ball will collide with the table at the time t=$t_c$ which satisfies
    $t_c$:=$h_i$-$\frac{1}{2}$g ${t_c}^2$=A cos(ω $t_c$-${\varphi }_i$)
This equation may not be solved in closed form.  However, we may solve it numerically (physical and mathematical arguments are available to convince us that we may always solve it).

Since the equation is completely elastic, the ball reflects up with a velocity
    $v_a$=-v($t_c$)+2V($t_c$)=g $t_c$-2A ω sin(ω $t_c$-${\varphi }_i$)
The ball will rise to height $h_f$ satisfying
    m g ($h_f$-$h_c$)=$\frac{1}{2}$m ${v_a}^2$
    $h_f$=A cos(ω $t_c$-${\varphi }_i$)+$\frac{1}{2g}$${v_a}^2$
    $h_f$=A cos(ω $t_c$-${\varphi }_i$)+$\frac{1}{2g}$${\left(g{t}_c-2A\omega \sin \left(\omega t_c-{\varphi }_i\right)\right)}^2$
at time
    g($t_f$-$t_c$)=$v_a$
    $t_f$=$\frac{1}{g}$$v_a$+$t_c$
    $t_f$=2$t_c$-$\frac{2A\omega }{g}$ sin(ω $t_c$-${\varphi }_i$)
At this time, the table will have phase
    ${\varphi }_f$=ω $t_f$-${\varphi }_i$=2ω $t_c$-$\frac{2A{\omega }^2}{g}$ sin(ω $t_c$-${\varphi }_i$)-${\varphi }_i$
    
That's it! That's the map! The deal is this.  You specify a system: choose your (ω, g, A).  Then, tell me a starting point: ($h_o$, ${\varphi }_o$).  From this, I may find $t_c$:
    $t_c$:=$h_i$-$\frac{1}{2}$g ${t_c}^2$=A cos(ω $t_c$-φ)
and then find $h_f$, ${\varphi }_f$:
    (

A pathological case

In[91]:=

$\mathrm{ho}=6.5;g=4.9;\omega =2;$

In[93]:=

$\mathrm{InterceptFindRoot}\left[\text{tt\_\_}\right]:=t/.\mathrm{FindRoot}\left[\mathrm{Cos}\left[2\pi \omega t\right]⩵\mathrm{ho}-\frac{1}{2}g{t}^2,\left\{t,\mathrm{tt}\right\}\right];$

In[94]:=

Out[94]=

$1.498328404055961$

Out[95]=

$1.618384425299805$

Out[96]=

$1.7496342988790012$

In[97]:=

$<<\text{Graphics`Graphics`}$

In[98]:=

$\mathrm{ho}=6.5;g=4.9;\omega =2;$

In[99]:=

$\mathrm{DisplayTogether}\left[\left\{\text{}\mathrm{Plot}\left[\left\{\mathrm{Cos}\left[2\pi \omega t\right],\mathrm{ho}-\frac{1}{2}g{t}^2\right\},\left\{t,0,2\right\}\right],\text{}\mathrm{ListPlot}\left[\left\{\left\{\mathrm{t1},\mathrm{Cos}\left[2\pi \omega \mathrm{t1}\right]\right\},\left\{\mathrm{t2},\mathrm{Cos}\left[2\pi \omega \mathrm{t2}\right]\right\},\left\{\mathrm{t3},\mathrm{Cos}\left[2\pi \omega \mathrm{t3}\right]\right\}\right\},\mathrm{PlotStyle}\to \mathrm{PointSize}\left[0.02\right]\right]\text{}\right\},\mathrm{PlotRange}\to \left\{-1.1,2\right\},\mathrm{ImageSize}\to 480\right];$

A perniciously pathological case

In[100]:=

$\mathrm{ho}=6.6;g=4.65;\omega =2;$

In[101]:=

$\mathrm{tbad}=\mathrm{InterceptFindRoot}\left[1.5\right]$

$\mathrm{FindRoot}\text{::}\mathrm{lstol}:\text{The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than}\mathrm{MachinePrecision}\text{digits of working precision to meet these tolerances.}\mathrm{More\dots }$

Out[101]=

$1.5485571342481255$

In[102]:=

$\mathrm{tgood}=\mathrm{InterceptFindRoot}\left[1.5,2\right]$

Out[102]=

$1.791845372744573$

In[103]:=

$\mathrm{DisplayTogether}\left[\left\{\text{}\mathrm{Plot}\left[\left\{\mathrm{Cos}\left[2\pi \omega t\right],\mathrm{ho}-\frac{1}{2}g{t}^2\right\},\left\{t,0,2\right\}\right],\text{}\mathrm{ListPlot}\left[\left\{\left\{\mathrm{tbad},\mathrm{Cos}\left[2\pi \omega \mathrm{tbad}\right]\right\},\left\{\mathrm{tgood},\mathrm{Cos}\left[2\pi \omega \mathrm{tgood}\right]\right\}\right\},\mathrm{PlotStyle}\to \mathrm{PointSize}\left[0.02\right]\right]\text{}\right\},\mathrm{PlotRange}\to \left\{-1.1,2\right\},\mathrm{ImageSize}\to 480\right];$

Simplify the collision time equation.

In order to fix this complication, we realize that once the ball crosses h=A, it must collide with the table in that period of oscillation.  So we can zoom in on just that period, and describe the position (phase) and velocity the ball enters that period with.  Then we will find a map to the phase and velocity of the next collsion. (We disregard absolute phase; if the ball skips two or five periods we don't care, we just want the phase it enters at next. That is to say, we map the phase onto the torus).

Let's non-dimensionalize and rescale the problem.  The following is horrible but mechanical; we are just trying to get a handle on the parameters of the new problem.

We drop a ball from a height $h_o$>A, and consider the time when the ball crosses h=A:
    h($t_A$)=A
    $h_o$-$\frac{1}{2}$g ${t_A}^2$=A
    $t_A$=$\sqrt{2\left(h_o-A\right)/g}$
so that
    h(t-$t_A$)=A-g $t_A$(t-$t_A$)-$\frac{1}{2}$g ${\left(t-t_A\right)}^2$
    v(t-$t_A$)=-g $t_A$-g (t-$t_A$)

The collision must occur within that period of the table's oscillation (because the ball drops continuously and the table will return to H=A at the end of the period).  Therefore, let us use
    ${\tau }_o$= [(ω $t_A$-φ) div 2π ]
    δ=(ω $t_A$-φ) mod  2π=ω $t_A$-φ-${\tau }_o$
    2 π τ=ω t-φ-2 π ${\tau }_o$
(div denotes integer division). Our motion is then
    H(t)=A cos(ω t-φ)=A cos(2 π τ+2 π ${\tau }_o$)=A Cos[2 π τ]    (${\tau }_o$ is an integer)
    h(t-$t_A$)=h($\frac{\tau -\delta }{\omega }$)=A-$\frac{g{t}_A}{\omega }$(τ-δ)-$\frac{g}{2{\omega }^2}$(τ-δ)
    v(t-$t_A$)=v($\frac{\tau -\delta }{\omega }$)=-g $t_A$-$\frac{g}{2\pi \omega }$ (τ-δ)

Finally, we set
    F(τ)=$\frac{H\left(t\right)}{A}$=Cos[2 π τ]
    W(τ)=$\frac{V\left(t\right)}{A2\pi \omega }$=-Sin[2 π τ]
    f(τ-δ)=$\frac{1}{A}$h(t-$t_A$)=1-β(τ-δ)-$\frac{\gamma }{2}$${\left(\tau -\delta \right)}^2$
    w(τ)=$\frac{1}{2\pi A\omega }$v(t)=-$\frac{g{t}_A}{A2\pi \omega }$-$\frac{g}{{A\left(2\pi \omega \right)}^2}$ (τ-δ)=-β-γ (τ-δ)
where
    β=$\frac{g{t}_A}{A2\pi \omega }$
    γ=$\frac{g}{A{\left(2\pi \omega \right)}^2}$.

So we see that we have one-D family of problems: you choose a γ, and then give me an initial velocity and phase (${\beta }_n$, ${\delta }_n$).  I can tell you the new velocity and phase (${\beta }_{n+1}$,${\delta }_{n+1}$).  Note that we have not restricted the problem at all; although the original problem let you pick an (A, ω, g), I could rescale by time and space and get an isomorphic problem.  We have reduced the number of parameters by removing this freedom.

The map, following an approach similar to that above, is to set a γ; then, using (${\beta }_n$,${\delta }_n$), find
    ${\tau }_c$:={1-β(${\tau }_c$-${\delta }_n$)-$\frac{\gamma }{2}$${\left({\tau }_c-{\delta }_n\right)}^2$=Cos[2 π ${\tau }_c$]}
    $w_c$:=-w(${\tau }_c$)+2W(${\tau }_c$)=β+γ (${\tau }_c$-δ)-2 Sin[2 π ${\tau }_c$]    (The motion right after the collision)

Now the motion after the collision is
    $w_a$(τ-${\tau }_c$)=$w_c$-γ(τ-${\tau }_c$)                (The motion at any time after the collision).
    $f_a$(τ-${\tau }_c$)=Cos[2 π ${\tau }_c$]+$w_c$(τ-${\tau }_c$)-$\frac{\gamma }{2}$${\left(\tau -{\tau }_c\right)}^2$
and the ball re-enters when
    Cos[2 π ${\tau }_c$]+$w_c$(${\tau }_f$-${\tau }_c$)-$\frac{\gamma }{2}$${\left({\tau }_f-{\tau }_c\right)}^2$==1
    (${\tau }_f$-${\tau }_c$)=$\frac{w_c+\sqrt{{w_c}^2-2\gamma \left(1-\mathrm{Cos}\left[2\pi {\tau }_c\right]\right)}}{\gamma }$        (take the positive solution to get re-entry)
    ${\tau }_f$=${\tau }_c$+$\frac{1}{\gamma }$($w_c$+$\sqrt{{w_c}^2-2\gamma \left(1-\mathrm{Cos}\left[2\pi {\tau }_c\right]\right)}$)
so the new phase is
    ${\delta }_{n+1}$=[${\tau }_c$+$\frac{1}{\gamma }$($w_c$+$\sqrt{{w_c}^2-2\gamma \left(1-\mathrm{Cos}\left[2\pi {\tau }_c\right]\right)}$)]mod 1

And the new velocity is
    ${\beta }_{n+1}$=-$w_a$(${\tau }_f$)=γ(τ-${\tau }_c$)-$w_c$=$\sqrt{{w_c}^2-2\gamma \left(1-\mathrm{Cos}\left[2\pi {\tau }_c\right]\right)}$

Giving the new map as
    ${\tau }_c$:={1-β(${\tau }_c$-${\delta }_n$)-$\frac{\gamma }{2}$${\left({\tau }_c-{\delta }_n\right)}^2$=Cos[2 π ${\tau }_c$]}
    $w_c$:=β+γ (${\tau }_c$-δ)-2 Sin[2 π ${\tau }_c$]
    (

Define the exact map.

In[104]:=

$\text{(*}\mathrm{An}\mathrm{annoying}\mathrm{Error}\mathrm{Message}\text{*)}h=\mathrm{Off}\left[\mathrm{FindRoot}\text{::}\mathrm{frsec}\right]$

In[105]:=

$\text{(*}\mathrm{The}\mathrm{collision}\mathrm{times}\mathrm{and}\mathrm{velocities}\text{*)}\mathrm{Find\tau c}\left[\left\{\text{δ\_},\text{β\_}\right\},\text{γ\_}\right]:=\tau /.\mathrm{FindRoot}\left[\mathrm{Cos}\left[2\pi \tau \right]==1-\beta \left(\tau -\delta \right)-\frac{\gamma }{2}{\left(\tau -\delta \right)}^2,\left\{\tau ,\delta ,1\right\}\right]\mathrm{ExactNewVel}\left[\left\{\text{δ\_},\text{β\_}\right\},\text{γ\_},\text{α\_},\text{τc\_}\right]:=\alpha \beta +\alpha \gamma \left(\mathrm{\tau c}-\delta \right)-\left(1+\alpha \right)\mathrm{Sin}\left[2\pi \mathrm{\tau c}\right]\text{(*}\mathrm{Assume}\alpha =1\mathrm{unless}\mathrm{otherwise}.\text{*)}\mathrm{ExactNewVel}\left[\left\{\text{δ\_},\text{β\_}\right\},\text{γ\_},\text{τc\_}\right]:=\mathrm{ExactNewVel}\left[\left\{\delta ,\beta \right\},\gamma ,1,\mathrm{\tau c}\right]\mathrm{ExactMap}\left[\left\{\text{δ\_},\text{β\_}\right\},\text{γ\_}\right]:=\mathrm{ExactMap}\left[\left\{\delta ,\beta \right\},\gamma ,1\right]\text{(*}\mathrm{The}\mathrm{exact}\mathrm{map}\text{*)}\mathrm{ExactMap}\left[\left\{\text{δ\_},\text{β\_}\right\},\text{γ\_},\text{α\_}\right]:=\mathrm{Block}\left[\left\{\mathrm{\tau c}:=\mathrm{Find\tau c}\left[\left\{\delta ,\beta \right\},\gamma \right],\mathrm{wc}:=\mathrm{ExactNewVel}\left[\left\{\delta ,\beta \right\},\gamma ,\alpha ,\mathrm{\tau c}\right],\mathrm{\beta new}:=\sqrt{{\mathrm{wc}}^2-2\gamma \left(1-\mathrm{Cos}\left[2\pi \mathrm{\tau c}\right]\right)},\mathrm{\tau new}:=\mathrm{Mod}\left[\delta +\frac{1}{\gamma }\left(\mathrm{wc}+\mathrm{\beta new}\right),1\right]\right\},\left\{\mathrm{\tau new},\mathrm{\beta new}\right\}\right]\text{(*}\mathrm{The}\mathrm{two}\mathrm{parts}\mathrm{of}\mathrm{the}\mathrm{motion}\text{*)}\mathrm{hbefore}\left[\left\{\text{δ\_},\text{β\_}\right\},\text{γ\_},\text{τ\_}\right]:=1-\beta \left(\tau -\delta \right)-\frac{\gamma }{2}{\left(\tau -\delta \right)}^2\mathrm{hafter}\left[\left\{\text{δ\_},\text{β\_}\right\},\text{γ\_},\text{τc\_},\text{wc\_},\text{τ\_}\right]:=\mathrm{Cos}\left[2\pi \mathrm{\tau c}\right]+\mathrm{wc}\left(\tau -\mathrm{\tau c}\right)-\frac{\gamma }{2}{\left(\tau -\mathrm{\tau c}\right)}^2$

A Pretty picture showing the ball bounce.

In[112]:=

$\mathrm{PlotPretty}\left[\text{δ\_},\text{β\_},\text{γ\_}\right]:=\mathrm{Module}\left[\left\{\right\},\{\text{(*}\mathrm{Find}\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{interest}\text{*)}\mathrm{\tau c}=\mathrm{Find\tau c}\left[\left\{\delta ,\beta \right\},\gamma \right],\mathrm{wc}=\mathrm{ExactNewVel}\left[\left\{\delta ,\beta \right\},\gamma ,\mathrm{\tau c}\right],\left\{\mathrm{\delta new},\mathrm{\beta new}\right\}=\mathrm{ExactMap}\left[\left\{\delta ,\beta \right\},\gamma \right],\mathrm{\tau a}=\mathrm{\tau c}+\frac{1}{\gamma }\left(\mathrm{wc}+\mathrm{\beta new}\right),\text{(*}\mathrm{Plot}\mathrm{the}\mathrm{motion}\text{*)}\mathrm{p1}:=\mathrm{Plot}\left[\left\{\mathrm{hbefore}\left[\left\{\delta ,\beta \right\},\gamma ,\tau \right],\mathrm{hafter}\left[\left\{\delta ,\beta \right\},\gamma ,\mathrm{\tau c},\mathrm{wc},\tau \right],\mathrm{Cos}\left[2\pi \tau \right],1\right\},\left\{\tau ,0,2\right\},\mathrm{DisplayFunction}->\mathrm{Identity}\right];,\text{(*}\mathrm{Plot}\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{interest}\text{*)}\mathrm{p2}:=\mathrm{ListPlot}\left[\left\{\left\{\delta ,\mathrm{hbefore}\left[\left\{\delta ,\beta \right\},\gamma ,\delta \right]\right\},\left\{\delta ,0\right\},\left\{\mathrm{\tau c},\mathrm{hbefore}\left[\left\{\delta ,\beta \right\},\gamma ,\mathrm{\tau c}\right]\right\},\left\{\mathrm{\tau c},0\right\},\left\{\mathrm{\tau a},\mathrm{hafter}\left[\left\{\delta ,\beta \right\},\gamma ,\mathrm{\tau c},\mathrm{wc},\mathrm{\tau a}\right]\right\},\left\{\mathrm{\tau a},0\right\}\right\},\mathrm{DisplayFunction}->\mathrm{Identity}\right];,\text{(*}\mathrm{Plot}\mathrm{the}\mathrm{graphs}\text{*)}\mathrm{Show}\left[\mathrm{p1},\mathrm{p2},\mathrm{PlotRange}->\left\{-1,3\right\},\mathrm{Prolog}->\mathrm{AbsolutePointSize}\left[4\right],\text{}\mathrm{ImageSize}\to 480,\text{}\mathrm{DisplayFunction}\to \mathrm{\$DisplayFunction}\right];\}\right];\text{(*}\mathrm{Block}\mathrm{PlotPretty}\text{*)}$

In[113]:=

$\text{(*}\mathrm{Some}\mathrm{representative}\mathrm{values}\text{*)}\mathrm{PlotPretty}\left[\text{.3},3,15\right]$

Out[113]=

$\left\{0.6200800327510121,9.171027669201457,\left\{0.28998997851412667,5.678822008510445\right\},1.610070011265139,\mathrm{Null},\mathrm{Null},\mathrm{Null}\right\}$

In[114]:=

$\mathrm{PlotPretty}\left[\text{.1},5,8\right]$

Out[114]=

$\left\{0.3850675179771662,5.9585529683476475,\left\{0.18710209724126003,2.7382638095824325\right\},1.472169615218426,\mathrm{Null},\mathrm{Null},\mathrm{Null}\right\}$

An approximation: Assume the table is stationary but delivers an impulse which depends on time (β is large)

Consider the case where β is large.  For instance,

In[115]:=

$\text{(*}A\mathrm{large}\beta \text{*)}\mathrm{PlotPretty}\left[\text{.5},15,30\right]$

Out[115]=

$\left\{0.6072555620575625,19.46575345415153,\left\{0.6986383758092645,16.493397820126408\right\},1.805893937866827,\mathrm{Null},\mathrm{Null},\mathrm{Null}\right\}$

In this case, the time between when the ball crosses h=A and when it collides with the table is small.  If this time is sufficiently small, the table will not move much in this time, and the ball will not accelerate much. For instance, in the graph above, the velocity of the ball and table at δ is about that at ${\tau }_c$.  

If we approximate ${\tau }_c$-${\delta }_n$=0,
    $w_c$:=${\beta }_n$+γ (${\tau }_c$-${\delta }_n$)-2Sin[2 π ${\tau }_c$]=${\beta }_n$-2Sin[2 π ${\delta }_n$]    (The ball doesn't accelerate much)
    (

Play with the approximate map.

In[116]:=

$\text{(*}\mathrm{The}\mathrm{first}\mathrm{approximate}\mathrm{map}\text{*)}\mathrm{ApproxMap}\left[\left\{\text{δ\_},\text{β\_}\right\},\text{γ\_}\right]:=\mathrm{Block}\left[\left\{\mathrm{\beta new}:=\beta -\mathrm{Sin}\left[2\pi \delta \right],\mathrm{\delta new}:=\mathrm{Mod}\left[\delta +\frac{1}{\gamma }\mathrm{\beta new},1\right]\right\},\left\{\mathrm{\delta new},\mathrm{\beta new}\right\}\right]$

In[117]:=

$\mathrm{NRandomSeeds}\left[\text{b\_},\text{range\_},\text{n\_}\right]:=\mathrm{Table}\left[\left\{\mathrm{Random}\left[\right],b+\mathrm{range}\mathrm{Random}\left[\right]\right\},\left\{i,1,n\right\}\right]$

In[143]:=

$\text{(*}\mathrm{NestList}\mathrm{gives}a\mathrm{list}\mathrm{of}\mathrm{the}\mathrm{results}\mathrm{of}\mathrm{applying}\mathrm{our}\mathrm{map}\mathrm{to}\mathrm{our}\mathrm{starting}\mathrm{point}\mathrm{to}\mathrm{expr}0\mathrm{through}n\mathrm{times}\text{*)}\mathrm{ApproxOrbit}\left[\left\{\text{δ0\_},\text{β0\_}\right\},\text{γ\_},\text{Iters\_}\right]:=\mathrm{NestList}\left[\mathrm{ApproxMap}\left[\text{\#},\gamma \right]\&,\left\{\mathrm{\delta 0},\mathrm{\beta 0}\right\},\mathrm{Iters}\right]\mathrm{ListApproxOrbits}\left[\text{Orbits\_},\text{Iters\_},\text{γ\_},\text{b\_},\text{range\_}\right]:=\left(\mathrm{ApproxOrbit}\left[\text{\#},\gamma ,\mathrm{Iters}-1\right]\&/@\mathrm{NRandomSeeds}\left[b,\mathrm{range},\mathrm{Orbits}\right]\right)/.\left(\left\{\text{x\_},\text{y\_}\right\}/;\left(x>0.5\right)\to \left\{x-1,y\right\}\right)\mathrm{PlotApproxOrbits}\left[\text{Orbits\_},\text{Iters\_},\text{γ\_},\text{b\_},\text{range\_}\right]:=\mathrm{ListPlot}\left[\mathrm{Flatten}\left[\mathrm{ListApproxOrbits}\left[\mathrm{Orbits},\mathrm{Iters},\gamma ,b,\mathrm{range}\right],1\right],\text{}\mathrm{ImageSize}\to 480\right];\mathrm{ThisApproxOrbit}\left[\text{Xo\_},\text{Iters\_},\text{γ\_}\right]:=\left(\mathrm{ApproxOrbit}\left[\mathrm{Xo},\gamma ,\mathrm{Iters}-1\right]\right)/.\left(\left\{\text{x\_},\text{y\_}\right\}/;\left(x>0.5\right)\to \left\{x-1,y\right\}\right)\mathrm{PlotThisApproxOrbit}\left[\text{Xo\_},\text{Iters\_},\text{γ\_}\right]:=\mathrm{ListPlot}\left[\mathrm{ThisApproxOrbit}\left[\mathrm{Xo},\mathrm{Iters},\gamma \right],\text{}\mathrm{ImageSize}\to 480\right];\text{(*}\mathrm{Prolog}->\left\{\mathrm{AbsolutePointSize}\left[4\right]\right\}\text{*)}$

A Plot of eight hundred points in the orbit of forty different initial (phase, velocity)s.

$\mathrm{PlotApproxOrbits}\left[40,800,6,8,12\right]$

A Plot of a few hundred points in the orbit of forty different initial (phase, velocity)s.

$\mathrm{PlotApproxOrbits}\left[40,300,6,9,5\right]$

A Couple orbits

In[148]:=

$\mathrm{PlotThisApproxOrbit}\left[\left\{\text{.25},12\right\},1000,6\right]$

Out[148]=

$⁃\mathrm{Graphics}⁃$

In[149]:=

$\mathrm{PlotThisApproxOrbit}\left[\left\{\text{.2},10.1\right\},1000,6\right]$

Out[149]=

$⁃\mathrm{Graphics}⁃$

In[150]:=

$\mathrm{PlotThisApproxOrbit}\left[\left\{\text{.32},12\right\},1000,6\right]$

Out[150]=

$⁃\mathrm{Graphics}⁃$

Here we notice periodic orbits and bounded regions.  There are several stable regions, correspoding to collsions with the peak of the table excursion.  This is because when the ball hits a little short of the peak, it gains some energy and so its next bounce is a little afterwards; then before, then after, and so on.  The circles around β=12 and β=18 correspond to stable regions in which 2, or 3, periods elapse between landings.  The map is repeated self-similarly every γ n in β.  

Too far from the peak, however, and you can hit a crisis:  bouncing higher can make you land where you get more energy, which can make you get more, and which can pop you into a different period between hops, which can pop you back.  That is to say, chaos.  The fuzzy regions (which are like the separatrices of the pendulum map) represent this behavior.

Within the fuzzyregions, however, there are islands of stability.  Each such island is self-similar to the full map.

It makes sense that the velocity is bounded.  Basically, the ball is equally likely to bounce in a trough (where the collision leaves the ball travelling more slowly) as in a peak (where the collision will speed up the ball).  In general, the average change in energy for each bounce is zero.

The three different limits for this problem.

This is not true for the exact map.  The ball is more likely to hit the table on the upstroke than on the downstroke.  Before the ball can hit the table, it must catch up to the table:

In[128]:=

$\mathrm{ho}=6.8;g=4.9;\omega =2;$

In[129]:=

$\mathrm{t4}=\mathrm{InterceptFindRoot}\left[1.5,2\right]$

Out[129]=

$1.7775069759013473$

In[130]:=

$\mathrm{DisplayTogether}\left[\left\{\text{}\mathrm{Plot}\left[\left\{\mathrm{Cos}\left[2\pi \omega t\right],\mathrm{ho}-\frac{1}{2}g{t}^2\right\},\left\{t,0,2\right\}\right],\text{}\mathrm{ListPlot}\left[\left\{\left\{\mathrm{t4},\mathrm{Cos}\left[2\pi \omega \mathrm{t4}\right]\right\}\right\},\text{}\mathrm{PlotStyle}\to \mathrm{PointSize}\left[0.02\right]\right]\text{}\right\},\mathrm{PlotRange}\to \left\{-1.1,2\right\},\mathrm{ImageSize}\to 480\right];$

The table is moving down, as is the ball; so the ball may not meet the table until it starts to rise again.  On average, the ball hits the table when it is moving upwards.  Therefore, each collsion on average pumps energy into the ball.  Eventually, the transit time between landings get so huge that the motion is chaotic: if the transit time is 40,000 oscillations, slightly different velocities can lead to very different landing phases.  Therefore, although the velocity will get large enough that the large β approximation is good, by that point things are too hairy to be interesting.

On the other hand, if we make the collision inelastic enough, we can make it so that the ball will lose velocity on each bounce. This leads to chattering: the ball will hit the table on the downstroke, lose some energy, hit on the down stroke twice more, then hit when the table is moving up, then again, gaining some energy but not too much... just enough that on the next collision the table loses energy again; and so on.  Then the ball will bounce along, making only small excursions from the table's position.  The main thing here is that all orbits eventually settle into this pattern, as long as velocity leaves the ball on each bounce.

Now we see why the approximate map is so popular:  by the way that we defined it, the ball is equally likely to hit any section of the table's oscillation.  In other words, on the average the ball's velocity remains the same.  In fact, this is the real reason the exact map is not pursued: its pictures are nowhere near as pretty.

I referred to papers by
    On the origin of the Cosmic Radiation.  Fermi, Phys Rev 75-8, p1169 (1949)
    Fermi Accelleration Revisited, Lichtenberg and Lieberman, Physica 1D 1980 p291
    Regular and Chaotic dynamics of the damped Fermi accelerator.  Luna-Acosta, Phys Rev. A         42-12 p 7155 (1990)
    Bouncing Ball with finite restitution: Chattering, Locking, Chaos.  Luck and Mehta, Phys. Rev. E         48-5, p. 3988 (1991)
    The Completely Inelastic Bouncing Ball.Luck and Mehta, Phys Rev Letters 65-4, p393 (Jul 1990)

Created by Mathematica  (June 3, 2007)