Question

When leaving from the port station at sea level, the Carmelit funicular travels 5,658 feet horizontally along a slope of 14.6/94.3. What is the elevation of the top station?

Answer 1

Answer: 876 feet

Step-by-step explanation:

Given

Carmelit travels 5658 feet horizontally

The slope of trajectory is   $\dfrac{14.6}{94.3}$

We know, the slope is

$\Rightarrow \text{slope}=\frac{\text{Vertical distance}}{\text{horizontal distance}}\\\\\Rightarrow \dfrac{14.6}{94.3}=\dfrac{Y}{5658}\\\\\Rightarrow Y=876\ feet$

Hence, the elevation of the top station is 876 feet