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 = square root of 3h/2, h>0 with d being the distance they can see in miles and h being their eye-level height in feet.

Answer 1

Using the formula d = √(3h/2), we find that Wyatt can see approximately 13.42 miles to the horizon and Shawn can see approximately 20.12 miles. Thus, Shawn can see 6.70 miles farther to the horizon than Wyatt.

Step-by-step explanation:

To calculate how much farther Shawn can see to the horizon than Wyatt, we apply the given formula d = √(3h/2), where d is the distance they can see in miles and h is their eye-level height in feet above sea level.

First, calculate the distance Wyatt can see using his height of 120 ft: d_Wyatt = √(3×120/2).

Then, calculate the distance Shawn can see using his height of 270 ft: d_Shawn = √(3×270/2).

Subtract Wyatt's distance from Shawn's to find the difference in the distances they can see to the horizon.

Lets calculate this step by step:

For Wyatt: d_Wyatt = √(3×120/2) ≈ √(180) ≈ 13.42 miles.

For Shawn: d_Shawn = √(3×270/2) ≈ √(405) ≈ 20.12 miles.

The difference: Shawn's horizon ≈ 20.12 miles − 13.42 miles ≈ 6.70 miles farther than Wyatt's.

Therefore, Shawn can see approximately 6.70 miles farther to the horizon than Wyatt can.

Answer 2

The answer is 6.7 miles further.

It is given:

`d= \sqrt{ (3h)/(2)}`
Wyatt’s eye-level height - h₁ = 120ft
Shawn’s eye-level height - h₂ = 270ft

The distance that Wyatt can see:

`d_1= \sqrt{ (3h_1)/(2)} = \sqrt{ (3*120)/(2 )}= √(180) =13.42miles`

The distance that Shawn can see:

`d_2= \sqrt{ (3h_2)/(2)}= \sqrt{ (3*270)/(2) }= √(405)=20.12 miles`

The question is how much farther can Shawn see to the horizon, so we need to subtract the results:
20.12 - 13.42 = 6.7 miles