Volume of regular tetrahedron of given length of edges

$y = \sqrt{3} ~ \text{ft}$
 

$h^2 + \left( \frac{2}{3}y \right)^2 = 2^2$

$h^2 + \left( \frac{2}{3}\sqrt{3} \right)^2 = 2^2$

$h = \sqrt{8/3} ~ \text{ft}$
 

Note that the base area is an equilateral triangle
$V = \frac{1}{3}A_b h$

$V = \frac{1}{3} \left[ \frac{1}{2}(2^2)\sin 60^\circ \right]\sqrt{\frac{8}{3}}$

$V = 0.9428 ~ \text{ft}^3$   ←   answer