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$