Answer:
The string of the kite is being let out at the rate of 19.2 ft/sec.
Step-by-step explanation:
See the diagram attached.
The vertical height (BK) of the kite is 120 ft.
The kite string is of length (AK) of 200 ft.
Therefore, the horizontal distance (AB) from Andrew to kite is = [tex]\sqrt{200^{2} - 120^{2}} = 160[/tex] ft.
Now, applying the Pythagoras Theorem,
AB² + BK² = AK²
⇒ l² + h² = s² {Where, l is the horizontal length, h is the height and s is the string length}
Differentiating with respect to time t both sides
[tex]2l\frac{dl}{dt} + 0 = 2s\frac{ds}{dt}[/tex] {Since, height of kite is constant}
⇒ [tex]2 \times 160 \times 24 = 2 \times 200 \times \frac{ds}{dt}[/tex]
⇒[tex]\frac{ds}{dt} = 19.2[/tex] ft per sec.
Therefore, when the wind is carrying the kite horizontally at the rate of 24 ft/sec, then the string of the kite is being let out at the rate of 19.2 ft/sec. (Answer)
