A kite is \(120 \mathrm{~m}\) high and \(130 \mathrm{~m}\) of string is out. If the kite is moving away horizontally at the rate of \(39 \mathrm{~m} / \mathrm{sec}\), then the rate at which the string is being out, is

Let 'P' be the position of the kite and PR be the string.
Let \(\mathrm{QR}=x\) and \(\mathrm{PR}=y\)
By Pythagoras theorem,
\(\begin{aligned}
& -\mathrm{PR}^2=\mathrm{PQ}^2+\mathrm{QR}^2 \\
& \Rightarrow y^2=(120)^2+x^2... (i)
\end{aligned}\)
Differentiating w.r.t.t, we get
\(\begin{aligned}
& 2 y \frac{\mathrm{d} y}{\mathrm{dt}}=2 x \frac{\mathrm{d} x}{\mathrm{dt}} \\
& \Rightarrow y \frac{\mathrm{d} y}{\mathrm{dt}}=x \frac{\mathrm{d} x}{\mathrm{dt}}... (ii)
\end{aligned}\)
Now, kite is moving away horizontally at the rate of \(39 \mathrm{~m} / \mathrm{sec}\).
\(\begin{aligned}
\therefore \quad & \frac{\mathrm{d} x}{\mathrm{dt}}=39 \mathrm{~m} / \mathrm{sec} \\
& \text { From (i), } \\
& (130)^2=(120)^2+x^2 \\
\Rightarrow & x^2=16900-14400 \\
\Rightarrow & x^2=2500 \\
\Rightarrow & x=50 \\
& \text { From (ii), } \\
& 130 \frac{\mathrm{d} y}{\mathrm{dt}}=50 \times 39 \\
\therefore \quad & \frac{\mathrm{d} y}{\mathrm{dt}}=\frac{50 \times 39}{130}=15 \mathrm{~m} / \mathrm{sec}
\end{aligned}\)