Center of mass

For a solid cone or pyramid, the centroid is $\frac{1}{4} \times$ the distance from the base to the apex.

Since the pyramid was $150m$ high, the centre of mass is $\frac{150}{4}m$ high.

$\Rightarrow$ The centre of mass is $\boxed{37.5}m$ high from the ground.

I recognise that there is a standard result I could have just googled but I'm learning a bunch of mechanics stuff at the moment about moments of inertia and there is a lot of similarity in the calculus treatment that can be given to the problem. I find it quite satisfying ;) I do apologise for the lack of diagrams but I have no software to work with here, I will do my best to describe!

First, symmetry tells us that the centre of mass is on the altitude line from peak to base of the pyramid, so that eliminates two dimensions and leaves us with just the height dimension. Imagine cutting the pyramid into infinitesimally thin slices, in a plane parallel to the base - then it becomes a problem of finding the centre of mass of an infinite number of points (since the centre of mass of each slice is the intersection of the two diagonals, which lies on the pyramid's altitude line), which is solved by integration.

Take a cross-section straight down the middle of the pyramid, in a plane perpendicular to the base that intersects the base parallel to one of the base edges. This is a triangle with height $h=150$ and base length $a=230$ . Laying this on an x-y coordinate system with the peak at the origin, and its altitude on the x-axis, we obtain the function $|y|=\frac{a}{2h}x$ . The bottom part is unnecessary, so just consider $y=\frac{a}{2h}x$

Remembering that the side-length of an infinitesimal slice through the pyramid is $2y$ on this graph, we can find the volume $\delta V$ of the slice, thickness $dx$ :

$\delta V = (2y)^{2}dx = \frac{a^{2}x^{2}}{h^{2}}dx$

The total volume of the pyramid could be taken as the integral over $x \in [0,h]$ , but the standard result is easily remembered as $V = \frac{1}{3}a^{2}h$ . Then the volume density, $\rho$ , of the pyramid is found:

$\rho = \frac{M}{V} = \frac{3M}{a^{2}}h$

The mass of our infinitesimal slice is calculated:

$\delta m = \rho \delta V = \frac{3Mx^{2}}{h^{3}}dx$

The perpendicular distance of the centre of mass from the base is denoted by $\overline{x}$ . Since for convenience the top of the pyramid is at the origin, observe the adjustment in the centre of mass formula:

$(h-\overline{x})M = \int_{0}^{h}x \delta m = \int_{0}^{h} \frac{3Mx^{3}}{h^{3}} dx= \frac{3}{4}Mh$

Solving obtains $\overline{x} = \frac{1}{4}h$ , and substituting the value of $h$ gives the final answer of $\overline{x}=\boxed{37.5}m$