I need more ropes

Classical Mechanics Level 3

I want to take the water from the well with a bucket attach to rope. The problem is, I don’t know the deep of the well, so I decided to drop a marble from the top of the well and 35 17 \frac{35}{17} second later the sound of the marble I dropped touched the water on the well heard. If g = 10 m/s 2 g = 10 \text{ m/s}^2 and the speed of sound is 340 m/s 340 \text{ m/s} , calculate the minimum length of the rope that I need to get the water from the well.


The answer is 20.

Muhammad Erfan by Muhammad Erfan , 30, Indonesia

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2 solutions

Adrian Peasey
Jul 24, 2015

Total time T T = time to fall t 1 t_1 + time for the sound to get back up t 2 t_2

Use SUVAT to find the time to fall the depth s s : s = u t + 1 2 a t 2 ⇒ s = 0 × t 1 + 1 2 × 10 × t 1 2 ⇒ t 1 = s 5 s=ut+\frac{1}{2}at^2\Rightarrow s=0\times t_1+\frac{1}{2}\times 10 \times t_1^2 \Rightarrow t_1=\sqrt{\frac{s}{5}}

To find the time to return: s = u t + 1 2 a t 2 ⇒ s = 340 × t 2 + 1 2 × 0 × t 2 ⇒ t 2 = s 340 s=ut+\frac{1}{2}at^2\Rightarrow s=340\times t_2+\frac{1}{2}\times 0\times t^2\Rightarrow t_2=\frac{s}{340}

This means we now have: T = t 1 + t 2 ⇒ 35 17 = s 5 + s 340 ⇒ ( s ) 2 + 68 5 s − 700 = 0 T=t_1+t_2 \Rightarrow \frac{35}{17}=\sqrt{\frac{s}{5}}+\frac{s}{340} \Rightarrow \left(\sqrt{s}\right)^2+68\sqrt{5}\sqrt{s}-700=0

Using the quadratic formula: x = − b ± b 2 − 4 a c 2 a ⇒ s = − 68 5 ± ( 68 5 ) 2 − ( 4 × 1 × − 700 ) 2 = − 68 5 ± 72 5 2 x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\Rightarrow \sqrt{s}=\frac{-68\sqrt{5}\pm\sqrt{\left(68\sqrt{5}\right)^2-\left(4\times1\times-700\right)}}{2}=\frac{-68\sqrt{5}\pm72\sqrt{5}}{2}

Since s > 0 \sqrt{s}>0 we have: s = − 68 5 + 72 5 2 = 2 5 = 20 \sqrt{s}=\frac{-68\sqrt{5}+72\sqrt{5}}{2}=2\sqrt{5}=\sqrt{20}

So the depth of the well and hence length of rope required is: ( s ) 2 = ( 20 ) 2 = 20 \left(\sqrt{s}\right)^2=\left(\sqrt{20}\right)^2=\boxed{20}

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By equating the the distance traveled during free fall and distance traveled by sound, we could get quadratic eqn. which could be solved easily using factorization.

0.t+gt^2/2=340(35/17-t)

,where t is the time traveled during free fall.

Zahle Khan - 5 years, 10 months ago

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Nice, Zahle. That's the trick to doing this one quickly.

Daniel Walvin - 5 years, 9 months ago

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Thank You .

Zahle Khan - 5 years, 9 months ago
Lu Chee Ket
Oct 10, 2015

s = u t + (1/2) a t^2 = v t

=> s = 0 + (1/2)(10)(35/17 - t)^2 = 340 t {Solving quadratic equation.}

=> t = 1/17 or 1225/17

But since time taken for sound has to be less than time taken by the drop, t = 1/ 17 only.

Hence s = 5 x 2^2 or 340/17 = 20 (meters)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I think this is a nice question.

Lu Chee Ket - 5 years, 8 months ago

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