A really, really good bouncy ball

H =initial height Hn =height after nth collision v=initial velocity just before collision vn=velocity after nth collision

time taken to hit the ground for first time is $\sqrt{2H/g}$ .

then tn=time taken for it to complete nth cycle up and down $=2(\sqrt{2Hn/g})$ .

let n collisions be have occurred by 1 hour

$3600= \sqrt{2H/g} + 2( \sqrt{2H1g} + \sqrt{2H2/g} +......+ \sqrt{2Hn/g}$

also

$vn=(1-e)^n v$

$Hn=((vn)^2)/2g$

or $Hn=((1-e)^2n) H$

using sum of GP for n terms and solving using calculator

n=2694 (nearest integer)

hence

$Hn=((0.9999)^{2694})10=7.6398$

Immediately before the first time that the ball hits the ground, all the initial potential energy will stay converted into kinetic energy. Let $v_0$ be the velocity of the ball immediatelly before the first hit. Then:

$\dfrac{mv_0^2}{2}=mgh_0$ $v_0=\sqrt{2\cdot9.8\cdot10}=14\ m/s$

Let $v_n$ be the velocity of the ball immediatelly after the $n$ -th hit (wich is equal in modulus to the velocity of the ball immediately before the $n+1$ -th hit). It is easy to see that:

$v_n=v_0\cdot (1-e)^n=14\cdot 0.9999^n\ m/s$

Lets calculate the time between the $n$ -th and the $n+1$ -th hit. Let $t_n$ be this time. Evidently, the ball will start and finish its trajectory at $h=0$ and its initial velocity will be $v_n$ . Then:

$0=v_nt_n-\dfrac{9.8t_n^2}{2}$ $9.8t_n^2-2v_nt_n=0$ $t_n(9.8t_n-2v_n)=0$

We are interested in the non-zero solution, so:

$9.8t_n-2v_n=0$ $t_n=\dfrac{2v_n}{9.8}$ $t_n=\dfrac{28}{9.8}\cdot 0.9999^n\ s$

Let $t_0$ be the time that the ball takes to first hit the ground. As the ball is intially at rest at $h=10\ m$ :

$0=10-\dfrac{9.8t_0^2}{2}$ $t_0=\sqrt{\dfrac{20}{9.8}}\ s$

Let $T_n$ be the time from the initial position to the $n$ -th hit. Then:

$T_n=t_0+\displaystyle\sum_i=1^n t_n=\sqrt\dfrac{20}{9.8}+\dfrac{28\cdot 0.9999}{9.8}\cdot\dfrac{1-0.9999^n}{1-0.9999}$

Letting $T_n=1\ h=3600\ s$ , we will obtain aproximatelly $n=1346.25$ . It means that the ball hits the ground $1346$ times after $1$ hour and is going to the $1347$ -th hit. So, we need to discover the height that the ball will reach when it hits the ground at the $1347$ -th time. The kinectic energy of the ball immediately after the (\1347)-th hit is, being $m$ its mass:

$\dfrac{m\cdot v_{1347}}{2}=m\cdot\dfrac{(14\cdot 0.9999^{1347})^2}{2}$

The ball will reach the maximum height when all the kinectic energy is transformed into potential energy. So:

$m\cdot\dfrac{(14\cdot 0.9999^{1347})^2}{2}=m\cdot 9.8\cdot h_{max}$ $h_{max}=\dfrac{(14\cdot 0.9999^{1347})^2}{2\cdot9.8}$ $h_{max}=7.63\ m$

From basic kinematics, the time for the ball to reach the ground from its initial height is $t_{0} = \sqrt{ \frac{2H}{g}}$ . The velocity $v_{0}$ with which it first impacts is $v_{0} = \sqrt{2Hg}$ . After it impacts, it rebounds with velocity $v = v_{0}(0.9999)^{n}$ , where n denotes the nth bounce. The time $t_{n}$ for each one of these subsequent bounces (up and down) is $t_{n} = \frac{2v}{g} = \frac {2(0.9999)^{n}v_{0}}{g}$ . The total time T to get to the Nth bounce is then $T = t_{0} + \sum_{n=1}^{N} t_{n}$ . Using our previous formulas, this becomes $T = \sqrt{ \frac{2H}{g}} + \sum_{n=1}^{N} \frac {2(0.9999)^{n}\sqrt{2Hg}}{g}$ .

The second term is just a finite geometric series, and can be expressed in closed form using the general formula $\sum_{k=1}^{j-1}a^{k} = a \left( {\frac {1-{r}^{j}}{1-r}}-1 \right)$ .

In our case $a = 2\sqrt{\frac{2H}{g}}, r = 0.9999, n = j + 1,$ and we get

$T = 2\,\sqrt {{\frac {2H}{g}}} \left( {\frac {1-{r}^{N+1}}{1-r}}-1\right)$ .

We want to find how many bounces occur in one hour, so we set T = 3600 seconds and solve for N. The algebra is tedious, and after some manipulation we get $N = \left( -\ln \left( r \right) +\ln \left( 1/4\, \left( 2\,\sqrt {2} \sqrt {{\frac {H}{g}}}r-T+Tr \right) \sqrt {2}{\frac {1}{\sqrt {{ \frac {H}{g}}}}} \right) \right) \left( \ln \left( r \right) \right) ^{-1}$

This expression evaluates to about 1346.8. Hence the ball is on its 1346th bounce after an hour. At this point it has $v \approx (0.9999)^{1346} \sqrt{2Hg} \approx 12.24$ m/s. With this velocity, the ball reaches height $h = \frac{v^{2}}{2g}$
So, $h \approx \frac{(12.24)^{2}}{2(9.8)} \approx 7.64$ meters.

First define $h_n$ as the height after the $n$ th bounce and $t_n$ as the time for the ball to go from $h_n$ to $h_{n+1}$ . Using the basic equations $\Delta h = v_o t - \frac{1}{2} g t^2$ and $v_f^2 - v_o^2 = - 2 g \Delta h$ , one obtains the relations

$\displaystyle t_n = \sqrt{\frac{2}{g}} (\sqrt{h_n} + \sqrt{h_{n+1}})$ and $h_{n+1} = v_{n+1}^2 / 2 g$ . From the problem, it is known that $v_{n+1} = (1 - e) v_n$ . Combining these equations together yields a recursion relation of $h_{n+1} = (1 - e)^2 h_n$ . This can be used to simplify $t_n$ to $\displaystyle t_n = (2 - e) \sqrt{\frac{2 h_n}{g}}$ .

Additionally $h_n$ can be put into closed form after defining $h_0$ to be the starting height: $h_n = h_0 (1 - e)^{2n}$ . Hence, $t_n$ can also be put into a closed form of $\displaystyle t_n = (2 - e) \sqrt{\frac{2 h_0}{g}} (1 - e)^n$ . The expression for approximate total time elapsed after $n$ bounces is then $\displaystyle T_n = \sum_{m=0}^n t_m$ . This is just a truncated geometric series and some simplification results in $\displaystyle T_n = \frac{2 - e}{e} \sqrt{\frac{2 h_0}{g}} (1 - (1 - e)^{n+1})$

Setting $T_n$ equal to 3600, solving for $n$ and plugging that back into $h_n$ yields an approximate height of 7.63 m

$\frac{\mathrm{d} v}{\mathrm{d} t}=-\frac{\epsilon v}{\Delta t}=-\frac{\epsilon v}{2 v} g=-\frac{\epsilon g}{2} \\ v(t)=v_0-\frac{\epsilon g t}{2}=\sqrt{2 g H}-\frac{epsilon g t}{2} \\ h=\frac{\left(\sqrt{2 g H}-\frac{epsilon g t}{2}\right)^2}{2 g}$

I have to call your attention about the time which passed since I submit my solution up to now when it was confirmed true. Agonizing for hours to find my mistake hehehe as at the beginning it was not accepted as correct.

The ball falls from 10 meters, so using 0.5gt^2=10 gives t=1.428 seconds for the time reaching to the ground. And it arrives at a speed v=gt=1.428xg=14m/sec. On the first bounce the ball will have a speed of14x0.9999, and on the second 14x(0.9999^2) and so on until bounce number n 14x(0.9999^n). The time the ball spends in the air after each bounce equals 2xVb/g, where Vb is the bouncing speed and g for gravity, and have to multiply by two for up and down trajectory. The speed series the ball makes looks: 14x0.9999,14x0.9999^.....,14X0.9999^n which is a geometric series.

The total time T the ball spends from it's release to n bounces is:

T=1.428+(2X14/g)x0.9999((1-0.9999^n)/ (1-0.999)) Where the total time, is given by the sum of the bouncing time geometric series added to the initial drop time. Now equating to T=3600 seconds for a one hour event, we get from above 0.9999^n=0.8740 where n stands for the number of bounces taking place in one hour. There is no need to solve for "n", as we know already that the speed after n bounces will be Vn=14x(0.9999^n)=14x0.874=12.24m/sec, from which we can compute the last bouncing height "h" according to the energy Eq. ((1/2)xmxVn^2=mxgxh)....

h=(Vn^2)/(2xg)=12.24^2/(2x9.8)=7.638=approx(7.64meter)

The relation between the velocity of the ball when it hits the ground and its maximum height $mv^2/2 = mgh$ ( $v^2 \sim h$ ), so $v \rightarrow (1-e)v \Rightarrow h \rightarrow (1-2e)h$ as $1 >> e$ .

After $n=1$ collision, $\Delta h = -2eh$ . The total time between two collisions is $T=\sqrt{\frac{8h}{g}}$ , so in the interval $dt$ we have $dn=dt/T=\sqrt{\frac{g}{8h}} dt$ collisions, so the height is decreasing with $dh = \Delta h dn = -e \sqrt{\frac{gh}{2}} dt$ :

$h^{-1/2} dh = -e \sqrt{\frac{g}{2}} dt \Rightarrow h^{1/2}-H^{1/2}=-e \sqrt{\frac{g}{8}} t \Rightarrow h(t) = (H^{1/2}-e \sqrt{\frac{g}{8}} t)^2 = 7.64(m)$

One can find the total time the ball bounces with:

$t=\frac1{e} \sqrt{\frac{8H}{g}} = 2.85 \times 10^4(s) = 7.94(h)$

Alternatively, one can just use the following equation:

$t=T(\frac12+(1-e) + (1-e)^2 + (1-e)^3 + ...) = (\frac{1}{e}-\frac12) T \approx \frac1{e} T = \frac1{e} \sqrt{\frac{8H}{g}}$

The time $\Delta t_{i}$ for an object with initial velocity v upwards to move a certain distance upward and the same distance downward (= one cycle) is given by :

$\Delta t_{i} = \dfrac{2v_{i}}{g}.$

For a total time t of bouncing, we thus find the following expression :

$t = \sum_{i=1}^{n} \left( \dfrac{2v_{i}}{g} \right) - \dfrac{v_{1}}{g},$

where the last term originates from the fact that the ball falls from a height of 10m, so it only completes half a cycle. I choose however, for reasons that will shortly follow to incorporate the full cycle in the sum.

Because of the loss in velocity during the bounce, we can write the velocity in the i-th step as

$v_{i} = e^{i-1} . v_{1} ,$

where $v_{1}$ is the velocity just before the ball bounces a first time. This gives us a geometric series for $v_{i}$ . If we know $v_{1}$ , we can easily calculate $v_{i}$ , so using conservation of energy we find :

$v_{1} = \sqrt{2 |g| H} = 14 \frac{m}{s}.$

Using t = 3600s, we find :

$17647 = \sum_{i=1}^{n} \left( v_{i} \right) = v_{1} . \dfrac{1-e^{n}}{1-e},$

wehere we used the fact that $v_{i}$ forms a geometric sequence, to our advantage.

We find $e^{n} = 0,87395$ , which we solve using logarithms finding $n \approx 1347$ .

This gives us for the height after one hour of bouncing :

$h_{1347} = \dfrac{g \Delta t_{i}^2}{2} = \dfrac{v_{i}^2}{2|g|} = \dfrac{(e^{1346} . v_{1})^2}{2|g|} \approx 7.6398 m$ .