Question

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

Answer 1

Given :
Height at which kite is flying $\left(h\right)=1515m$
Speed of kite $\left(V\right)=10m/s$

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

$\Rightarrow \frac{dx}{dt}=10$

From the figure, we have
$EC=151.5-15=150m$

Let $AE=x\text{and}AC=y$

In right angle $△CEA$.
$A{E}^2+E{C}^2=A{C}^2$

[Using Pythagoras theorem]

$\Rightarrow x^2+(150{)}^2=y^2\cdots (1)$

Differentiating both sides w.r.t. $t$, we get

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

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

$\Rightarrow \frac{dy}{dt}=\frac{x}{y}\frac{dx}{dt}\cdots \left(2\right)$

When $y=250m,$

$\Rightarrow x^2+(150{)}^2=(250{)}^2$ [From eq. (1)]

$\Rightarrow x=200$

${\left(\frac{dy}{dx}\right)}_{y=250}=\frac{200}{250}.10=8m/s$

$[\because \frac{dx}{dt}=10m/s]$

Hence, the required rate at which the string is being let out is $8m/s$