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# 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

Toolbox:
• If $y = f(x)$, then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
• $\bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
Given height at which the kite is flying is
$AC=151.5 m$
Height of the boy = $BC= 1.5m$
$\Rightarrow\:AB=h=151.5-1.5=150$
By applying pythagoras theorem in $\Delta \:ABE$,
$AB^2+BE^2=AE^2$
$i.e.,\:\:\:h^2+x^2=y^2\: \: \: \: \: \: \: \: \: \: \: (1)$
when $x=BE=250\:m$,
$150^2+250^2=AE^2$
$\Rightarrow AE=y=291.5m$
Differentiating equation (1) w.r.t. $t$ we get
$2x.\large\frac{dx}{dt}$$=2y.\large\frac{dy}{dt} \Rightarrow x\large\frac{dx}{dt}=y \large\frac{dy}{dt} ( Since height is always a constant ) Step 2 Given \large\frac{dx}{dt}$$=10m/sec$ and $x=250\:m$
Now substituting the values for $x, y$ and $\large\frac{dx}{dt}$ we get
$10 \times 250 = 291.5 \times \large\frac{dy}{dt}$
$\Rightarrow \large\frac{dy}{dt}= \large\frac{10 \times 250}{291.5}$
$= 8 m/sec$
Hence the rate at which the string is released is $8m/ sec$
edited Jan 18, 2014
Jitender the correction has been done. Please  go through