A kite is moving horizontally at a height of 151.5 m. If the speed of kite is 10 m/s, how fast is the string being let out, when the kite is 250 m away from the boy who is flying the kite, if the height of boy is 1.5 m?

8 m/s

The correct answer is Option (2) → 8 m/s ##

We have, height $(h) = 151.5 \text{ m}$, speed of kite $(v) = 10 \text{ m/s}$

Let $CD$ be the height of kite and $AB$ be the height of boy.

Let $DB = x \text{ m} = EA$ and $AC = 250 \text{ m}$

$∴\frac{dx}{dt} = 10 \text{ m/s}$

From the figure, we see that

$EC = CD - ED \Rightarrow CD - AB$ [$∵ED = AB$]

$EC = 151.5 - 1.5 = 150 \text{ m}$

and $AE = x$ [$∵DB = AE$]

Also, $AC = 250 \text{ m}$

In right angled $\triangle CEA$,

$AE^2 + EC^2 = AC^2$

$\Rightarrow x^2 + (150)^2 = y^2 \dots(i)$

$\Rightarrow x^2 + (150)^2 = (250)^2$

$\Rightarrow x^2 = (250)^2 - (150)^2$

$= (250 + 150)(250 - 150) = 400 \times 100$

$∴x = 20 \times 10 = 200$

From Eq. (i), on differentiating w.r.t. $t$, we get

$2x \cdot \frac{dx}{dt} + 0 = 2y \frac{dy}{dt}$

$\Rightarrow 2y \frac{dy}{dt} = 2x \frac{dx}{dt}$

$∴\frac{dy}{dt} = \frac{x}{y} \frac{dx}{dt}$

$= \frac{200}{250} \cdot 10 = 8 \text{ m/s}$  $[∵\frac{dx}{dt}=10\,m/s]$

So, the required rate at which the string is being let out is 8 m/s.