# finding the grade of a mountain trail

• Dec 29th 2009, 09:51 PM
alessandromangione
finding the grade of a mountain trail
a straight trail with a uniform inclination leads from a hotel, elevation 5000 feet, to a lake in a valley, elevation 4100 feet. The lenght of the trail is 4100 feet. what is the inclination ( grade) of the trail

i'm not a native speaker of english , tha's why i don't really understand...i hope u guys can help me
• Dec 30th 2009, 01:17 AM
Hello alessandromangione
Quote:

Originally Posted by alessandromangione
a straight trail with a uniform inclination leads from a hotel, elevation 5000 feet, to a lake in a valley, elevation 4100 feet. The lenght of the trail is 4100 feet. what is the inclination ( grade) of the trail

i'm not a native speaker of english , tha's why i don't really understand...i hope u guys can help me

The usual way of giving the grade of a hill is as a percentage. It is calculated from the formula:
Percentage grade $\displaystyle = \frac{\text{vertical distance}}{\text{horizontal distance}}\times 100$ %
If the incline is very small, this is approximately the same as:
Percentage grade $\displaystyle \approx \frac{\text{vertical distance}}{\text{distance along slope}}\times 100$ %
In the question you have been given, the vertical distance $\displaystyle = 5000-4100=900$, and the distance along the slope $\displaystyle = 4100$. This is quite a steep incline, so it's not a good idea to use the second formula. So we use Pythagoras' Theorem to work out the horizontal distance:
Horizontal distance $\displaystyle = \sqrt{4100^2-900^2}=4000$
So, using the first formula:
Percentage grade $\displaystyle = \frac{900}{4000}\times 100$ % $\displaystyle = 22.5$ %