Answer
a)
$P\left( h \right)=473{{e}^{-0.000048h}}$.
b)
$\text{243 lb/f}{{\text{t}}^{2}}$.
c)
$\text{100 lb/f}{{\text{t}}^{2}}$.
d)
$14,441\ \text{ft}$
Work Step by Step
a)
Consider the exponential decay function, $P\left( t \right)={{P}_{0}}{{e}^{-kt}}$ …… (1)
Here, the value of ${{P}_{0}}$ is $473\ \text{lb/f}{{\text{t}}^{2}}$.
$P\left( h \right)=473{{e}^{-kh}}$
Now, substitute, $82,345\ \text{ft}-36,152\ \text{ft}=46,193\text{ ft}$ for $h$ and $51\ \text{lb/f}{{\text{t}}^{2}}$ for $P\left( h \right)$ and calculate the value of $k$ as,
$\begin{align}
& P\left( h \right)=473{{e}^{-kh}} \\
& 51=473{{e}^{-k\left( 46193 \right)}} \\
& \frac{51}{473}={{e}^{-k\left( 46193 \right)}}
\end{align}$
$\begin{align}
& \ln \frac{51}{473}=\ln {{e}^{-k\left( 46193 \right)}} \\
& \ln \frac{51}{473}=-k\cdot 46193 \\
& \frac{\ln \frac{51}{473}}{46193}=-k \\
& 0.000048\approx k
\end{align}$
b)
Consider the above expression, $P\left( h \right)=473{{e}^{-0.000048h}}$
Here, the height is equal to $50,000-36,152=13,848\ \text{ft}$, so put $h=13,848\text{ ft}$ in the above equation as,
$\begin{align}
& P\left( h \right)=473{{e}^{-0.000048h}} \\
& P\left( 13,848 \right)=473{{e}^{-0.000048\left( 13,848 \right)}} \\
& P\left( 13,848 \right)=473{{e}^{-0.664707}} \\
& P\left( 13,848 \right)=473\left( 0.5144 \right)
\end{align}$
Further simplified,
$\begin{align}
& P\left( 13,848 \right)=473\left( 0.5144 \right) \\
& P\left( 13,848 \right)=243.3 \\
& P\left( 13,848 \right)\approx 243 \\
\end{align}$
c)
Consider the above expression, $P\left( h \right)=473{{e}^{-0.000048h}}$
We find:
$\begin{align}
& P\left( h \right)=473{{e}^{-0.000048h}} \\
& 100=473{{e}^{-0.000048h}} \\
& \frac{100}{473}={{e}^{-0.000048h}}
\end{align}$
$\begin{align}
& \ln \frac{100}{473}=\ln {{e}^{-0.000048h}} \\
& \ln \frac{100}{473}=-0.000048h \\
& \frac{\ln \frac{100}{473}}{-0.000048}=h \\
& 32,373\approx h
\end{align}$
d)
Consider the above expression, $P\left( h \right)=473{{e}^{-0.000048h}}$
Here, the atmospheric pressure is being halved, so put $P\left( h \right)=\frac{1}{2}\left( 473 \right)\text{ or 236}\text{.5}$ in the above equation and calculate the value of $h$:
$\begin{align}
& P\left( h \right)=473{{e}^{-0.000048h}} \\
& 236.5=473{{e}^{-0.000048h}} \\
& \frac{236.5}{473}={{e}^{-0.000048h}}
\end{align}$
$\begin{align}
& \ln \frac{236.5}{473}=\ln {{e}^{-0.000048h}} \\
& \ln \frac{236.5}{473}=-0.000048h \\
& \frac{\ln \frac{236.5}{473}}{-0.000048}=h \\
& 14,441\approx h
\end{align}$
