Bouncing stacked balls over Mt. Everest

Designate the lowest ball as $B_1$ , the highest ball as $B_n$

For simplicity, assume that every ball separated by a very small distance, so that the relevant bounces happen a short time apart. This assumption isn’t necessary, but it makes for a slightly cleaner solution.

At the precise moment when $B_1$ reaches the ground, all of the balls are moving downward with speed $v_0=\sqrt{2gh}$ . After that, $B_1$ will collide with $B_2$ , which will collide with $B_3$ , and so forth.

We can inductively determine the speed of each ball after it bounces off the one below it.
If $B_i$ achieves a speed of $v_i$ after bouncing off $B_{i-1}$ , then what is the speed of $B_{i+1}$ after it bounces off $B_i$ ?
The relative speed of $B_{i+1}$ and $B_i$ (right before they bounce) is $v_0 + v_i$ .
Taking into account the conservation of momentum and $B_i$ being significantly larger than $B_{i+1}$ , after colliding with $B_{i+1}$ , $B_i$ is still moving upwards at essentially speed $v_i$ , so the final upward speed of $B_{i+1}$ is $v_{i+1}=(v_0 + v_i) + v_i$ . Thus, we arrive at:

$v_{i+1}=v_0+2v_i$ where $v_1=v_0$ . This is a geometric series, so we can get :

$v_{n}=(2^{n}-1)\sqrt{2gh}$

From conservation of energy, all of the kinetic energy of $B_n$ will turn into potential energy at its highest point.
Thus, the highest point $B_n$ will bounce to is:

$H=\frac{((2^n-1)v)^2}{2g}=(2^n-1))^2h$

Note that $g$ does not appear in this equation so, the highest point will actually be the same on any planet.
If $H$ is larger than 8848, and we need $2^n-1>\sqrt{8848}$ . Thus arriving at the final answer: $n=7$ , which will bounce the ball to the height of $16129$ meters.

Of course, this could only happen in ideal conditions, 3 stacked balls would boost the top ball 49 times the initial height, but in the video, the golf ball only bounced to 8 times its original height (which is still amazing).

This answer is based on Havard Physics Undergraduate Problem Week 9/16/02
Watch the full video on youtube: Stacked ball drop

First, we make the simplifying assumption that all the dropped balls have gaps between them, so that only 2 balls are in contact at any given time. We examine what happens when 2 balls collide.

Given masses $m_1>>m_2$ , let's assume that they respectively have velocities $kv, -v$ , where $k$ is an integer and postive velocity is upwards. Then we know from both conservatiion of kinetic energy and momentum the following equations:

$\dfrac{1}{2}(m_1(kv)^2+m_2(-v)^2 = \dfrac{1}{2}(m_1(kv+v_1)^2+m_2(-v+v_2)^2$

$m_1(kv)+m_2(-v) = m_1(kv+v_1)+m_2(-v+v_2)$

Solving for the unknowns $v_1, v_2$ , we have

$v_1=-\dfrac{2(1+k)m_2v}{m_1+m_2}$

$v_2=\dfrac{2(1+k)m_1v}{m_1+m_2}$

In the limiting case where $m_1>>m_2$ , we have

$v_1=0$
$v_2=2(1+k)v$

which means the top ball bounces up at a speed

$v_2-v=(1+2k)v$

We need at least an upward speed of $\sqrt{\dfrac{8898}{1}}v \approx 94v$ to go higher than Mt Everest, so we look at the ratio for each successive ball, starting with just 1 ball

$(1+2\cdot0)v=1v$
$(1+2\cdot1)v=3v$
$(1+2\cdot3)v=7v$
$(1+2\cdot7)v=15v$
$(1+2\cdot15)v=31v$
$(1+2\cdot31)v=63v$
$(1+2\cdot63)v=127v$

so 7 balls are needed for the job

The reason why the video only shows an 800% gain in altitude with 3 balls is because the ratio of weight of successive balls isn't significantly large, and certainly not at the limiting case.