Question

wyatt’s eye - level height is 120 ft above sea level, and shawn’s eye - level height is 270 ft above sea level. how much farther can shawn see to the horizon? use the formula $d=sqrt{\frac{3h}{2}}$, $hgeq0$, with $d$ being the distance they can see in miles and $h$ being their eye - level height in feet.
$sqrt{5}$ mi
$3sqrt{5}$ mi
$15sqrt{6}$ mi
$45sqrt{5}$ mi

Explanation:

Step1: Calculate Shawn's distance

Substitute \( h = 270 \) into \( d=\sqrt{\frac{3h}{2}} \).
\( d_{Shawn}=\sqrt{\frac{3\times270}{2}}=\sqrt{405}=\sqrt{81\times5}=9\sqrt{5} \)

Step2: Calculate Wyatt's distance

Substitute \( h = 120 \) into \( d=\sqrt{\frac{3h}{2}} \).
\( d_{Wyatt}=\sqrt{\frac{3\times120}{2}}=\sqrt{180}=\sqrt{36\times5}=6\sqrt{5} \)

Step3: Find the difference

Subtract Wyatt's distance from Shawn's distance.
\( 9\sqrt{5}-6\sqrt{5}=3\sqrt{5} \)

Answer:

\( 3\sqrt{5} \) mi (corresponding to the option "3√5 mi")