For jets, how can we show that 1/2*ρ*V^2 is equivalent to 1481.3*δ*M^2?
Using the Mach number definition $M=\frac{V}{\sqrt{\gamma RT}}$, the ideal gas equation $p=\rho RT$ and the atmospheric pressure ratio $\delta=\frac{p}{p_0}$:

$$ \frac{1}{2} \rho V^2 = \frac{1}{2} \rho \cdot \gamma RT \frac{V^2}{\gamma RT} = \frac{1}{2} \gamma p \cdot M^2 = \frac{\gamma \cdot p_0}{2} \cdot \delta \cdot  M^2 $$

With $p_0=2116.2 \text{lbs}/\text{ft}^2$ and $\gamma = 1.4$ you get the desired result.