Question

The Cave of Swallows is a natural open-air pit cave in the state of San Luis Potosí, Mexico. The 1220-foot-deep cave was a popular destination for BASE jumpers. The function 1/4sqrt(d) represents the time t (in seconds) that it takes a BASE jumper to fall d feet. How far does a BASE jumper fall in 3 seconds? Pls answer this as quickly as possible. Thanks.

Answer 1

The depth to which a BASE jumper jumps in 3 seconds is 144 feet

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;

`t = (1)/(4) \cdot√(d)`

The distance a BASE jumper jumps in 3 seconds = Required

By substituting t = 3 in the given function, we get;

`t = 3 = (1)/(4) \cdot√(d)`

Therefore;

4 × 3 = 12 = √d

d = 12² = 144

The distance a BASE jumper jumps in 3 seconds is d = 144 feet.