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

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

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

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

Therefore, $\frac{{dx}}{{dt}} = 10\;{\rm{m}}/{\rm{s}}$

From the figure, we see that
$EC = 151.5 - 1.5 = 150\;{\rm{m}}$

and $AE = x$
Also $AC = 250\;{\rm{m}}$

In right angled $\Delta CEA$

$A{E^2} + E{C^2} = A{C^2}$

$\Rightarrow$ ${x^2} + {(150)^2} = {y^2}$ ……..(i)
$\Rightarrow$ ${x^2} + {(150)^2} = {(250)^2}$

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

$= (250 + 150)(250 - 150)$

$= 400 \times 100$

Therefore,$x = 20 \times 10 = 200$

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

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

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

Therefore,$\frac{{dy}}{{dt}} = \frac{x}{y} \cdot \frac{{dx}}{{dt}}$

$= \frac{{200}}{{250}} \cdot 10 = 8\;{\rm{m}}/{\rm{s}}$

So, the required rate at which the string is being let out is $8\;{\rm{m}}/{\rm{s}}$.