Question

300

A helicopter lifts off the ground vertically at a contant rate of 8 feet per second. An observer watches the
helicopter lift off the ground at a point 300 feet away from the ground directly below the helicopter.
Find the rate that the distance between the observer and the helicopter is changing, in feet per second, when the
helicopter is 125 feet above the ground.

Answer 1

The rate that the distance between the observer and the helicopter is changing is 10 feet per second.

Step-by-step explanation:

To find the rate that the distance between the observer and the helicopter is changing, we can use the concept of similar triangles.

Let's denote the distance between the observer and the helicopter as d, the height of the helicopter as h, and the rate of the helicopter as r.

We are given that the rate at which the helicopter is lifting off the ground is 8 feet per second, so r = 8.

Using the concept of similar triangles, we can set up the following proportion:

(300 - d) / d = (h - 125) / 125

Cross multiplying and simplifying, we get:

125(300 - d) = d(h - 125)

Now, we can substitute r = 8 and solve for d:

125(300 - d) = 8d(300 - 125)

Simplifying further, we get:

37500 - 125d = 2400d - 1000

Adding 125d to both sides and combining like terms, we have:

12500 = 1375d

Dividing both sides by 1375, we get:

d = 10

So, the distance between the observer and the helicopter is changing at a rate of 10 feet per second when the helicopter is 125 feet above the ground.