$A$ kite is moving horizontally at a height of $151.5 \ m$. If the speed of the 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? The height of the boy is $1.5 \ m$.

(8 M/S) Let the height of the kite be $CD = 151.5 \ m$ and the height of the boy be $AB = 1.5 \ m$. Let the horizontal distance between the boy and the kite be $x$ and the length of the string be $y$.
From the geometry of the problem,the effective height of the kite above the boy's eye level is $h = 151.5 - 1.5 = 150 \ m$.
Using the Pythagorean theorem in the right-angled triangle formed by the string,the horizontal distance,and the effective height:
$x^2 + h^2 = y^2$
$x^2 + (150)^2 = y^2$ --- $(i)$
Given that the kite moves horizontally at $10 \ m/s$,we have $\frac{dx}{dt} = 10 \ m/s$.
When the string length $y = 250 \ m$,we find $x$ from $(i)$:
$x^2 + 150^2 = 250^2$
$x^2 = 62500 - 22500 = 40000$
$x = 200 \ m$.
Differentiating $(i)$ with respect to $t$:
$2x \frac{dx}{dt} + 0 = 2y \frac{dy}{dt}$
$x \frac{dx}{dt} = y \frac{dy}{dt}$
Substituting the values:
$200 \times 10 = 250 \times \frac{dy}{dt}$
$\frac{dy}{dt} = \frac{2000}{250} = 8 \ m/s$.
Thus,the string is being let out at a rate of $8 \ m/s$.