A kite flies with a constant height of 60 meters above ground. The speed is $2 \frac ms$. How fast is the line released the moment $100$ meters of it is released?

My attempt

I visualized this as a right triangle, where the hypotenuse $L$ is the line, the horizontal edge is $x$ with $x' = 2\frac ms$ and the vertical edge is $60\mathrm m$.

I tried creating a function, $$x(L) = \sqrt{L^2 - 60^2}$$

which gives $$x'(L) = \frac{L}{\sqrt{L^2 - 60^2}}$$

From here I'm unsure what to do. What I want to know is $L'$ when $L = 100$ right?

This is where I'm stuck. Any help appreciated!


Note that the length of the horizontal edge as a function of time is $x(t)=2t$. Hence by the Pythagorean theorem: $$L(t)=\sqrt{60^{2}+x(t)^{2}}=\sqrt{60^{2}+4t^{2}}.$$ Let's see when $L(t)=100$: $$100=\sqrt{60^{2}+4t^{2}} \Leftrightarrow t=40.$$

And at $t=40$ the line is released with speed $$L'(40)=\frac{8\cdot 40}{2\sqrt{60^{2}+4\cdot40^{2}}}=1.6\ \frac{m}{s}.$$

  • $\begingroup$ Ah, see that's where I mislead myself. I should have treated each length as a function of time, rather than just have them as functions of each other. Thanks! $\endgroup$
    – Alec
    Oct 21 '16 at 11:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.