# Penny Dropped down Well Problem

• Fusilli_Jerry89
In summary: In this case, the 72.26 seconds would not make sense in the context of the problem. The only relevant solution is the 0.086 seconds, which is the time it takes for the sound to reach the top of the well. Therefore, in summary, by using the formula for distance and solving with the quadratic equation, we can determine that the well is approximately 28.5 meters deep. The time it takes for the sound to reach the top of the well is 0.086 seconds, while the time it takes for both the penny and the sound to reach the bottom of the well is 2.5 seconds. The negative solution obtained during the calculation is irrelevant and has no physical significance in this context.
Fusilli_Jerry89
A penny is dropped down a well. One hears the sound of it hitting the bottom 2.5 seconds later. If sounds travels at 330 m/s, how deep is the well.

What I did:
Time(penny)+Time(sound)=2.5
T(p)=2.5-T(s)
T(s)=x
T(p)=2.5-x
d=vit+1/2at^2
330x=1/2(-9.8)(2.5-x)^2
I then solved using the quadratic forumula and got -62.2 or-0.102. I then chose the 0.102 seconds and multipled it by 330 and got 33.7 m. If this is right, my question is how can both times be negative, and how can you decifer which one is right besides using common sense?

Fusilli_Jerry89 said:
A penny is dropped down a well. One hears the sound of it hitting the bottom 2.5 seconds later. If sounds travels at 330 m/s, how deep is the well.

What I did:
Time(penny)+Time(sound)=2.5
T(p)=2.5-T(s)
T(s)=x
T(p)=2.5-x
d=vit+1/2at^2
330x=1/2(-9.8)(2.5-x)^2
I then solved using the quadratic forumula and got -62.2 or-0.102. I then chose the 0.102 seconds and multipled it by 330 and got 33.7 m. If this is right, my question is how can both times be negative, and how can you decifer which one is right besides using common sense?
You have sign problems? If the penny fell for the full 2.5 seconds, how far could it fall?

no the time that it fell plus the time is took for you to hear it totals 2.5 seconds.

Fusilli_Jerry89 said:
no the time that it fell plus the time is took for you to hear it totals 2.5 seconds.
I know that. The point is that in 2.5 seconds the penny could not even fall as far as you said it did. You made a mistake in your calculations.

weel can you show me what i did wrong?

Fusilli_Jerry89 said:
weel can you show me what i did wrong?
I don't see all your work, but it is probably here that you went wrong
330x=1/2(-9.8)(2.5-x)^2
330x is the distance the sound moves, a positive number

1/2(-9.8)(2.5-x)^2 is supposed to be the distance the penny moves, another positive number, but you made it negative. My guess is that if you get rid of the - in fornt of the 9.8 it will solve your problem.

isnt acceleration always negative tho in regards to gravity? and I thought since the penny is moving down, isn't that a negative direction?

Last edited:
I also did what you said and got 28.5 m which seems right. But for the time it takes the sound to reach the top, i got 0.086 sec or 72.26 sec. Why do you get two answers in this case? What is the significance of the 72.26 seconds?

Fusilli_Jerry89 said:
I also did what you said and got 28.5 m which seems right. But for the time it takes the sound to reach the top, i got 0.086 sec or 72.26 sec. Why do you get two answers in this case? What is the significance of the 72.26 seconds?
This is common when you solve quadratic equations. You often get a solution that is mathematically correct, but has no physical significance.

## What is the "Penny Dropped down Well Problem"?

The "Penny Dropped down Well Problem" is a hypothetical physics problem that involves a penny being dropped down a well and the calculation of its speed and distance traveled before hitting the bottom.

## What are the key factors that affect the solution to this problem?

The key factors that affect the solution to this problem include the height of the well, the initial velocity of the penny, and the acceleration due to gravity.

## What is the equation used to solve this problem?

The equation used to solve the "Penny Dropped down Well Problem" is: d = v0*t + 1/2*g*t^2, where d is the distance traveled, v0 is the initial velocity, g is the acceleration due to gravity, and t is the time.

## How does the mass of the penny affect the solution?

The mass of the penny does not affect the solution to this problem, as it is cancelled out in the equation for distance traveled. However, it does affect the speed at which the penny falls and the force with which it hits the bottom of the well.

## What are some real-world applications of this problem?

This problem can be applied to various scenarios, such as calculating the speed and distance of an object dropped from a certain height, or determining the time it takes for an object to fall a certain distance. It can also be used in engineering and construction to calculate the force of impact and design safety measures accordingly.

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