models of motion: A stone is released from rest and....

mathstresser

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Jan 28, 2006
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I should know how to do this. But for some reason I don't.

A stone is released from rest and dropped into a deep well. 8 seconds later, the sound of the stone splashing into the water at the bottom of the well returns to the ear of the person who released the stone. How long does it take the stone to drop to the bottom of the well? How deep is the well? Ignore air resistance. Note: the speed of sound is 340 m/s.

Does that mean that it takes a total of 8 seconds to drop the stone and hear the splash? Or does it mean that after the stone hits the water, the splash can be heard 8 seconds later?

Either way, I don't know what to do.
 
8 seconds for the round trip.

Calculate the time to descend. You can use your standard acceleration formulation.

The time for the sound to travel is at a constant speed. That is the easy part.
 
I still don't know how to use the equation.

F=ma
That just leaves me with acceleration= gravity.

a=x''
x''=-g
x'= -gt + c = -gt + v0
x'= v
x= -.5 gt^2 + v0t +x0

I can't find t, though, because I don't know x (the position) because I don't know how deep the well is. What should I be doing differently?
 
mathstresser said:
x(t)= -.5 gt^2 + v0t +x0
We know \(\displaystyle V_{0} = 0 m/s\). Do you see why? What words give it away?

\(\displaystyle x(t) = -\frac{1}{2} g t^{2} + x_{0}\)

\(\displaystyle x_{0}\) is the value we seek.

Define \(\displaystyle t_{0}\) to be the time to drop.

\(\displaystyle \;-\frac{1}{2} g t_{0}^{2} + x_{0} = 0\)

\(\displaystyle x_{0} = \frac{1}{2} g t_{0}^{2}\)

\(\displaystyle \sqrt{\frac{2x_{0}}{g}} = t_{0}\) <== There's how long it takes to hit the water.

Define \(\displaystyle t_{1}\) to be the time for the sound to get back up.

\(\displaystyle t_{1}*340 m/s = x_{0}\) <== The same distance back up.

\(\displaystyle t_{1} = \frac{x_{0}}{340 m/s}\) <== There's how long it takes for you to hear it hit the water.

\(\displaystyle t_{0} + t_{1} = 8 sec\)

That's all the pieces.
 
Re: models of motion

Hello, mathstresser!

This is a messy one . . .


A stone is released from rest and dropped into a deep well.
8 seconds later, the sound of the stone splashing into the water at the bottom of the well
returns to the ear of the person who released the stone.

(a) How long does it take the stone to drop to the bottom of the well?
(b) How deep is the well? .(Note: the speed of sound is 340 m/s.)

Eight seconds is the total time: .the time for the stone to hit the water
. . and for the sound to reach the person's ear.

The free-fall equation for a dropped object is: \(\displaystyle \:y \:=\:h\,-\,4.9t^2\)
. . where \(\displaystyle h\) is the original height
. . and \(\displaystyle y\) is the height of the stone at time \(\displaystyle t\).

When does the stone hit the water? .When \(\displaystyle y \,=\,0\)
. . We have: \(\displaystyle h\,-\,4.9t^2\:=\:0\;\;\Rightarrow\;\;t\:=\;\sqrt{\frac{h}{4.9}}\) . . . this is \(\displaystyle T_1\)

How long does it take for the sound to reach the ear?
. . \(\displaystyle T_2\:=\:\frac{h}{340}\)

The total time is 8 seconds: \(\displaystyle \:T_1\,+\,T_2\:=\:\sqrt{\frac{h}{4.9}}\,+\,\frac{h}{340}\;=\;8\)


We have: \(\displaystyle \:\sqrt{\frac{h}{4.9}}\;=\;8\,-\,\frac{h}{340}\)

Square both sides: \(\displaystyle \:\frac{h}{4.9}\;=\;64\,-\,\frac{16}{340}h\,+\,\frac{h^2}{115,600}\)

Multiply by \(\displaystyle 4.9(340^2) \,=\,566,440:\;\;115,600h \;=\;36,252,160\,-\,26,656h\,+\,4.9h^2\)

We have the quadratic: \(\displaystyle \:4.9h^2\,-\,142,256h\,+\,36,252,160\;=\;0\)

Quadratic formula: \(\displaystyle \:h\;=\;\frac{142,256\,\pm\,\sqrt{152,356^2\,-\,4(4.9)(36,252,160)}}{2(4.9)} \;=\;\left\{\begin{array}{cc}28,774.72217 \\ 257.1145588\end{array}\)


We reject the larger answer. .(A well 28.8 km deep?)

Therefore: (b) the depth of the well is: \(\displaystyle \fbox{257.11\text{ m}}\)


And: (a) the time of falling is: .\(\displaystyle \sqrt{\frac{257.11}{4.9}} \:\approx\:\fbox{7.24\text{ seconds}}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

The sound reaches the ear in \(\displaystyle \,\frac{257.11}{340}\:\approx\:\fbox{0.76\text{ seconds}}\)

. . And there is your 8 seconds!

 
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