Question

Watching a Helicopter Take Off At a distance of 58 ft from the pad, a man observes a helicopter taking off from a heliport. If the helicopter lifts off vertically and is rising at a speed of 45 ft/sec when it is at an altitude of 124 ft, how fast is the distance between the helicopter and the man changing at that instant? (Round your answer to one decimal place.)

Answer 1

To solve this problem we will apply the concepts related to classical mechanics, for which we will first consider the general equation that describes the movement of the body. From there, as we well know, we will obtain the derivative, which is equivalent to the speed of the body and the rate of change that is investigated in the problem.

Our values are given as,

$h = 58ft$

$v = 45ft/s$

$h' = 124ft$

We will start by defining the general height 'y' of the body at a reference distance of the lifting copter, given as

$y^2 = (58)^2+(45t)^2$

$y^2 = 3364+2025t^2$

Differentiating both the sides we have

$2y \frac{dy}{dt} = 2025*2t$

$\frac{dy}{dt} = \frac{2025t}{y}$

Now when the altitude of the copter is 124ft we have that

$y^2 = (58)^2+(124)^2$

$y^2 = 18740$

$y = 136.89ft$

And,

$45t=124$

$t = 2.75s$

Using these values at the derivative equation finally we have that

$\frac{dy}{dt} = \frac{2025t}{y}$

$\frac{dy}{dt} = \frac{2025(2.75)}{136.89}$

$\frac{dy}{dt} = 40.68ft/s$

Therefore the distance between the man and the helicopter is increasing at the rate of 40.68ft/s