Solve Freefall Problem: Find Water's Depth in Well

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In summary, the rock falls from rest and splashs when it hits the water at the bottom of the well. The sound of the splash is heard 2.42 seconds later. Using a kinematic equation, we find that the rock's initial position is 0 and its velocity is 0 at the time of the sound. We then use the sound reaching you equation to find that the rock's X coordinate is (1/2) a (T1)^2 and its distance is (1/2) a (T2)^2. Finally, we combine these three equations to find that the rock falls 26.5 meters from the top of the well.
  • #1
Freeze
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Ok I am at this problem for hours and I am still nowhere, I am just wondering if any of you can help.

Here is the question

A rock is dropped from rest into a well. If the sound of the splash is heard 2.42s later, how far below the top of the well is the surface of the water? (take the speed of sound to be 336.0 m/s and take the freefall acceleration of the rock to be 9.8 m/s^2.

So I figure ok this must mean

2.42s = Time spent falling + Amount of time it took for the sound to reach the top.

So I try to use one of the kinematic equations for motions with constant acceleration

Sf = Si + Vi*(Change of T) + 1/2 * A *(Change of T) ^ 2

Since Initial Position Si = 0 and Vi = 0 The formula will look like

Sf = 1/2 *A*(Change of T) ^ 2

Since I need to get rid of one of the unknowns I do this

Sf = 1/2 * 9.8 * (2.42s - SF/336m/s) ^ 2

Sf = 4.9 * (2.42s- SF/336m/s) ^ 2

Sf = 4.9 * (813.12m - SF)/336m/s)^ 2

And then yeah..I went further then that but let's just say it didn't give me the right answer.

I tried several other methods but nothing gave me the right answer.

Any Ideas on how to resolve this matter?
 
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  • #2
It seems that you use only one change of time but there are two times t1 fall time, t2 is time for the sound to get back up. And t1+t2=2.42

For the distance we know that S=0,5g*t1 but also S=Vsound*t2 and
t2=2.42-t1

So now you can make youre square equation with standard solutions.
 
  • #3
Sorry, I have been out of school for one year, and I now have no idea how to make square equation with standard solutions.

Can you provide an example please?
 
  • #4
come on guys, I really need some help here.
 
  • #5
Here's how I would do it.
Let T1 = time for rock to fall
Let T2 = time for sound to reach you

We know from freefall :
X = (1/2) a T^2

so
X = (1/2) a (T1)^2

Then from the sound reaching you we have:
X = Velocity * Time

X = V*T2

combine that with first as X=X

V*T2 = (1/2) a * (T1)^2

We also know T1+T2 = 2.42 seconds
T2 = 2.42 - T1
so
336*(2.42-T1) = (1/2)*9.8*(T1)^2
Write as a quadratic equation :
(multiply things out, then get T1^2 alone, etc.

T1^2 + 68.6 T1 - 165.9 = 0

Use solution of quadratic equations to get Ti, then plub into x=(1/2)(9.8)T1^2 to get X. (search google if you don't know solution of quadratic).

thats how I would do it.
I get 26.5 meters.
 
  • #6
Wow, you gave a really easy to follow explanation and that answer matches with the answer from the back of the book as well.

Thank you very much.
 

Related to Solve Freefall Problem: Find Water's Depth in Well

1. How do you solve a freefall problem to find water's depth in a well?

The first step in solving a freefall problem to find water's depth in a well is to identify the known variables, such as the acceleration due to gravity (9.8 m/s^2) and the initial velocity (usually 0 m/s). Next, use the equation d = 1/2 * a * t^2 to calculate the time it takes for an object to reach the bottom of the well. Finally, use the equation d = v0 * t + 1/2 * a * t^2 to calculate the depth of the well, where v0 is the initial velocity and t is the time calculated in the previous step.

2. What is the acceleration due to gravity used in this problem?

The acceleration due to gravity used in this problem is typically 9.8 m/s^2. This value is based on the acceleration of objects due to the Earth's gravitational pull.

3. Can this problem be solved without knowing the initial velocity?

Yes, this problem can still be solved without knowing the initial velocity. In this case, the equation d = 1/2 * a * t^2 can be used to calculate the time it takes for an object to reach the bottom of the well. Then, the equation d = 1/2 * a * t^2 can be used to calculate the depth of the well, where t is the time calculated in the previous step.

4. Is it necessary to convert units when solving this problem?

Yes, it is necessary to convert units when solving this problem. The equations used to solve this problem require consistent units, so it is important to convert all values to either SI units (meters, seconds) or imperial units (feet, seconds) before plugging them into the equations.

5. Can this problem be solved using different equations?

Yes, there are multiple equations that can be used to solve this problem, such as the kinematic equations or the energy conservation equation. However, the equations mentioned in the previous questions are the most commonly used and are sufficient for solving this problem.

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