Answer:
The depth to which a BASE jumper jumps in 3 seconds is 144 feet
Step-by-step explanation:
The details of the Cave of Swallows are;
The depth of the cave = 1,220 ft.
The function that represents the duration, t, in seconds it takes to fall d feet is given as follows;
[tex]t = \dfrac{1}{4} \cdot\sqrt{d}[/tex]
The distance a BASE jumper jumps in 3 seconds = Required
By substituting t = 3 in the given function, we get;
[tex]t = 3 = \dfrac{1}{4} \cdot\sqrt{d}[/tex]
Therefore;
4 × 3 = 12 = √d
d = 12² = 144
The distance a BASE jumper jumps in 3 seconds is d = 144 feet.
