Thus the radius of ball is decreasing at a constant rate. We have, height $(\mathrm{h})=151.5 \mathrm{~m}$, speed of kite $(\mathrm{v})=10 \mathrm{~m} / \mathrm{s}$ Let $\mathrm{CE}$ be the height of kite and $\mathrm{AB}$ be the height of boy. Let $\mathrm{CB}=\mathrm{x} \mathrm{m}=\mathrm{DA}$ and $\mathrm{AE}=250 \mathrm{~m}$
$$
\therefore \frac{\mathrm{dx}}{\mathrm{dt}}=10 \mathrm{~m} / \mathrm{s}
$$


From the figure, we see that
$$
\mathrm{ED}=151.5-1.5=150 \mathrm{~m}
$$
and $\mathrm{AD}=\mathrm{x}$
Also, $\mathrm{AE}=250 \mathrm{~m}$
In right angled $\triangle \mathrm{DEA}$,
$$
\begin{aligned}
& \mathrm{AD}^2+\mathrm{ED}^2=\mathrm{AE}^2 \\
\Rightarrow \quad & \mathrm{x}^2+(150)^2=\mathrm{y}^2 \\
\Rightarrow \quad & \mathrm{x}^2+(150)^2=(250)^2 \\
\therefore \quad & \mathrm{x}=200
\end{aligned}
$$
From Eq. (i), on differentiating w.r.t.t. we get
$$
\begin{aligned}
&\therefore \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{x}}{\mathrm{y}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}} \\
&=\frac{200}{250} \cdot 10=8 \mathrm{~m} / \mathrm{s} \quad\left[\because \frac{\mathrm{dx}}{\mathrm{dt}}=10 \mathrm{~m} / \mathrm{s}\right]
\end{aligned}
$$
So, the required rate at which the string is being let out is $8 \mathrm{~m} / \mathrm{s}$