24. flying a kite once kates kite reaches a height of 50 ft (above her hands), it rises no higher but drifts…

Explanation:

Step1: Set up the Pythagorean theorem

Let ( y = 50) (constant height), (x) be the horizontal distance, and (s) be the length of the string. Then (s^{2}=x^{2}+y^{2}). Since (y = 50), we have (s^{2}=x^{2}+2500).

Step2: Differentiate with respect to time (t)

Differentiating both sides of (s^{2}=x^{2}+2500) with respect to (t): [ \begin{align*} 2s\frac{ds}{dt}&=2x\frac{dx}{dt}+0\ s\frac{ds}{dt}&=x\frac{dx}{dt} \end{align*} ] We know that (\frac{dx}{dt}=5) (horizontal speed). When (s = 120), we can find (x) using (s^{2}=x^{2}+y^{2}). So (x=\sqrt{s^{2}-y^{2}}=\sqrt{120^{2}-50^{2}}=\sqrt{14400 - 2500}=\sqrt{11900}=10\sqrt{119}).

Step3: Solve for (\frac{ds}{dt})

Substitute (s = 120), (x = 10\sqrt{119}), and (\frac{dx}{dt}=5) into (s\frac{ds}{dt}=x\frac{dx}{dt}): [ \begin{align*} 120\frac{ds}{dt}&=10\sqrt{119}\times5\ \frac{ds}{dt}&=\frac{50\sqrt{119}}{120}\ \frac{ds}{dt}&=\frac{5\sqrt{119}}{12}\approx4.47\space ft/s \end{align*} ]

Answer:

(\frac{5\sqrt{119}}{12}\approx4.47\space ft/s)