Hello, mistykz!

A girl flies a kite at a height of 300 ft,

the wind carrying the kite horizontally away from her at a rate of 25 ft/sec.

How fast must she let out the string when the kite is 500 ft away from her? I wouldn't introduce trigonometry . . .

Code:

B x C
* - - - - - - - - * →
| *
| *
300 | * S
| *
| *
A * - - - - - - - - - - - - -

The girl is at $\displaystyle A.$

The kite is at height of 300 feet: $\displaystyle AB = 300$

. . and is moving from point $\displaystyle C$ at the rate of $\displaystyle \frac{dx}{dt} = 25$ ft/sec.

The string has length: $\displaystyle S \,=\,AC$

Pythagorus says: .$\displaystyle S^2\:=\:x^2+300^2$ .**[1]**

Differentiate with respect to time: .$\displaystyle 2S\left(\frac{dS}{dt}\right) \:=\:2x\left(\frac{dx}{dt}\right) $

. . Hence, we have: .$\displaystyle \frac{dS}{dt} \:=\:\frac{x}{S}\cdot\frac{dx}{dt}$ .**[2]**

When $\displaystyle S = 500$, [1] becomes: .$\displaystyle 500^2 \:=\:x^2+300^2\quad\Rightarrow\quad x \:=\:400$

Substitute into [2]: .$\displaystyle \frac{dS}{dt} \:=\:\frac{400}{500}(25) \:=\:20$ ft/sec