College Algebra, Chapter 3, 3.1, Section 3.1, Problem 72

Due to the curvature of the Earth, the maximum distance $D$ that you can see from the top of a tall building or from an airplane at height $h$ is given by the function $D(h) = \sqrt{2rh + h^2}$, where $r = 3960$mi is the radius of the Earth and $D$ and $h$ are measured in miles.
a.) Find $D(0.1)$ and $D(0.2)$
For $D(0.1)$,

$\begin{equation} \begin{aligned} D(0.1) &= \sqrt{2(3960 \text{mi})(0.1 \text{mi}) + (0.1\text{mi})^2} && \text{Replace } h \text{ by } 0.1\\ \\ &= 28.14 \text{mi} \end{aligned} \end{equation}$


For $D(0.2)$,

$\begin{equation} \begin{aligned} D(0.2) &= \sqrt{2(3960 \text{mi})(0.2 \text{mi}) + (0.2\text{mi})^2} && \text{Replace } h \text{ by } 0.1\\ \\ &= 39.8 \text{mi} \end{aligned} \end{equation}$


b.) How far can you see from the observation deck of Toronto's CN Tower, 1135ft above the ground?

$\begin{equation} \begin{aligned} D(h) &= \sqrt{2rh + h^2}\\ \\ D(1135\text{ft}) &= \sqrt{22(3960 \text{mi})\left(1135\cancel{\text{ft}}\right) \left( \frac{1\text{mi}}{5280\cancel{\text{ft}}} \right) + \left[ \left(1135 \cancel{\text{ft}} \right) \left( \frac{1\text{mi}}{5280\cancel{\text{ft}}} \right) \right]^2 }\\ \\ &= 41.26 \text{mi} \end{aligned} \end{equation}$


You can see $41.26 \text{mi}$ from the observation deck of Toronto's CN Tower that is 1135ft above the ground.

c.) Commercial Aircraft fly at an altitude of about 7mi. How far can the pilot see?

$\begin{equation} \begin{aligned} D (h) &= \sqrt{2rh + h^2}\\ \\ D(7\text{mi}) &= \sqrt{2(3960 \text{mi})(7 \text{mi}) + (7\text{mi})^2}\\ \\ &= 235.56 \text{mi} \end{aligned} \end{equation}$


The pilot will see 235.56mi from an altitude of 7mi.