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

Explanation:

Given

Carmelit travels 5658 feet horizontally

The slope of trajectory is
`(14.6)/(94.3)`

We know, the slope is

`\Rightarrow \text{slope}=\frac{\text{Vertical distance}}{\text{horizontal distance}}\\\\\Rightarrow (14.6)/(94.3)=(Y)/(5658)\\\\\Rightarrow Y=876\ feet`

Hence, the elevation of the top station is 876 feet