Across a stationary normal shock, $p_2/p_1=1+[2\gamma/(\gamma+1)](M_1^2-1)$. Upstream air has speed 800 m/s, static temperature 300 K and pressure 1 atm. With γ=1.4 and R=287 J/(kg K), find downstream static pressure in atm (two decimal places).
$M_1^2=V_1^2/(\gamma RT_1)=640000/(1.4\times287\times300)=5.3094$. Substitute into the given shock relation: $p_2/p_1=1+(2.8/2.4)(5.3094-1)=6.0276$. Since p₁=1 atm, p₂=6.03 atm.
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