Go Stack Marbles on the Freeway!

This kind of stacking has octahedral symmetry, i.e., cab form an octahedron, which means that the side edges are $45°$ from the plane. We thus have a stack of $6$ balls tilted at that angle, and so we can work out how high this tiled stack of balls is from plane to top

$10(1+5\dfrac { 1 }{ \sqrt { 2 } } )=45.3553...$

I used Pythagorean Theorem.. x^2 +(25sqrt/{2})^2=50^2.. Solving for x, we get x=45.35533906..

In the stack of cannon balls,

The bottom corner cannon ball to top cannon ball will make 45 degree angle at their respective centers.

There are six layers

The distance from center of bottom corner cannon ball to top cannon ball center = 5(bottom corner cannon ball radius) + 40(diameter of 4 cannon balls in between) + 5(top cannon ball radius)

Let this distance be D, thus D=50 cm.

This 45 degree is subtended to height of the stack. Thus,

\sin { { 45 }^{ \circ } } = \frac {D}{50}

\frac { 1 }{ \sqrt { 2 } } =\frac { D }{ 50 } \ \ D\quad =\quad \frac { 50 }{ \sqrt { 2 } }

D = 35.36

Further 2 points to be noted:

1. We had considered solution from center, for the first layer cannon ball. So, the radius of the cannon ball is to be added to the height of the stack.

2. We had considered solution from center, for the top layer cannon ball. So, the raius of the cannon ball is to be added to the height of the stack.

Thus the total height of the Stack H = D + 5 + 5

H = 45.36 cm