A kite flying at a a height h meter has r meter of string paid out at a time of t sec. If the kite moves horizontally with constant velocity v meter/sec then the rate at which the string is paid out is

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$\frac{r \sqrt{(r^{2} - h^{2})}}{v} mt / \sec$

$\frac{\sqrt{r^{2} - h^{2}}}{rv} mt / \sec$

$\sqrt{(r^{2} - h^{2})} vmt / \sec$

$\frac{v \sqrt{(r^{2} - h^{2})}}{r} mt / \sec$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{dx}{dt} = v, y = r, x = \sqrt{r^{2} - h^{2}} \\ y^{2} = r^{2} = x^{2} + h^{2} \Rightarrow 2 y \frac{dy}{dt} = 2 x \frac{dx}{dt} \\ \Rightarrow \frac{dy}{dt} = \frac{x}{y} \frac{dx}{dt} = v \frac{\sqrt{r^{2} - h^{2}}}{r} \end{array}$

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