Chapter 1.2, Problem 100E

To determine

The maximum distance that one can see from the observation deck of the Toronto CN tower.

Answer to Problem 100E

The maximum distance that one can see from the observation deck of the Toronto CN tower is $\underset{\_}{2.18\times {10}^5\text{ft}}$.

Explanation of Solution

Formula used:

$D=\sqrt{2rh+h^2}$

Where D is the maximum distance that can be seen from the top of a tall building, h is the height of the building, r is the radius of the earth.

Calculation:

Given that, the radius of the earth is $3960\text{ mi}$ and the height of the observation deck of the Toronto CN Tower is $1135\text{ ft}$.

It is known that, $1\text{mi}=5280\text{ft}$.

Use the unit conversion to convert the distance $3960\text{mi}$ into feet as shown below.

$\begin{array}{c}3960\text{mi}=3960\times 1\text{mi}\\ \text{=3960}\times 5280\text{ft}\\ \text{=20908800}\text{ft}\end{array}$

Therefore, the radius of the earth is $\text{20908800}\text{ft}$.

Substitute $\text{20908800}\text{ft}$ for r, $1135\text{ft}$ for h in the formula to calculate D as follows.

$\begin{array}{c}D=\sqrt{2\times \left(20908800\text{ft}\right)\times \left(1135\text{ft}\right)+{\left(1135\text{ft}\right)}^2}\\ =\sqrt{47462976000+1288225}\text{ft}\\ =\sqrt{47464264225}\text{ft}\\ =217862.948\text{ft}\end{array}$

Now, move decimal 5 places to the left and write 5 as power of 10 to express the above quantity in proper scientific notation.

$\begin{array}{c}D=217862.948\text{ft}\\ =2.17862948\times {10}^5\text{ft}\end{array}$

Round off the above quantity up to 2 decimal places as shown below.

$\begin{array}{c}D=2.17862948\times {10}^5\text{ft}\\ \text{=2}\text{.18}\times {10}^5\text{ft}\end{array}$

Therefore, the maximum distance that one can see from the observation deck of the Toronto CN tower is $\underset{\_}{2.18\times {10}^5\text{ft}}$.