= Solution
\b[The <P wave> source is normally the more useful depth source for the end members and for the stated approximately 50% gas case], because the predicted $|R_p|$ exceeds $|R_s|$, making the reservoir-top arrival easier to identify, assuming comparable source and recording quality. For a horizontal reflector and vertical two-way <travel time> $t$ entirely through the homogeneous cap rock,
$$
h=\frac{V_{p1}t_p}{2}=750t_p\ \mathrm m,
$$
or $h=V_{s1}t_s/2=300t_s\ \mathrm m$ for an identified <S wave> arrival, with times in seconds. The reservoir's own <wave speed> is not the overburden <wave speed> used in this conversion. Nonzero-offset data require <moveout correction> and an appropriate overburden <velocity model>; no numerical depth can be obtained without a measured time.
At approximately 50% gas, Figure 1 gives $\rho\simeq1.5\ \mathrm{g\,cm^{-3}}$, $V_p\simeq1.05\ \mathrm{km\,s^{-1}}$, and $V_s\simeq0.62$–$0.63\ \mathrm{km\,s^{-1}}$. Thus
$$
Z_{p2}\simeq1.58\times10^6,\quad R_p\simeq-0.18,\qquad Z_{s2}\simeq(0.93\text{–}0.945)\times10^6,\quad R_s\simeq+0.02.
$$
The <S wave> reflector is then weak because its impedance is nearly matched to the cap rock. An exact <impedance match>, $\rho_2V_{s2}=\rho_1V_{s1}$, gives $R_s=0$: a reservoir top can exist without producing an observable normal-incidence <S wave> reflection. Rounding the 50% $V_s$ to $0.60$ would suggest exactly that cancellation, but the plotted line sits slightly higher. The continuous $Z_s$ curve crosses the cap-rock value at a somewhat greater saturation; the coarse figure does not justify a precise percentage.
There is also a smaller-saturation <P wave> impedance match: $R_p$ changes sign between zero and modest gas saturation, so it must pass through zero. Thus neither source guarantees a detectable reflector at every saturation. Joint <P wave> and <S wave> observations help distinguish a weak impedance contrast from an absent reservoir; a slower <S wave> speed alone is no reason to prefer it when its reflected amplitude is near zero.
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