A kite is flying at a height of 60 m. At the instant when 100 m of string is out, the kite is moving horizontally away from the person holding the string at the rate of 5 m/s. The rate at which the string is being let out at that instant is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4 m/s
Let x be the horizontal distance of the kite from the person and s the length of the string. The height stays 60 m, so by Pythagoras' theorem
\(s^{2} = x^{2} + 60^{2}\)
Differentiating with respect to time t (the rates of change are linked):
\( \displaystyle 2s\frac{ds}{dt} = 2x\frac{dx}{dt} \)
When s = 100:
\( \displaystyle x = \sqrt{100^{2} - 60^{2}} = \sqrt{6400} = 80 \)
With \(\frac{dx}{dt} = 5\):
\( \displaystyle \frac{ds}{dt} = \frac{x}{s} \cdot \frac{dx}{dt} = \frac{80}{100} \times 5 \)
\( = 4 \text{ m/s} \)
Why the other options are incorrect
Option (1) 3 m/s — Incorrect. This is \(\frac{60}{100} \times 5\): it uses the height 60 m in place of the horizontal distance 80 m.
Option (2) 5 m/s — Incorrect. The string is slanting, so it is let out more slowly than the kite moves horizontally: \(\frac{ds}{dt} = \frac{x}{s} \cdot \frac{dx}{dt}\) with \(\frac{x}{s} \lt 1\).
Option (3) 6.25 m/s — Incorrect. This is \(\frac{100}{80} \times 5\), the ratio \(\frac{x}{s}\) turned upside down.