# Solve Physics Question: Rock Dropped Down Mine Shaft

In summary, the student dropped a rock down a deep mineshaft and heard it hit the bottom 6.5 seconds later. To find the depth of the mineshaft, the student can use the equation h=343*t, where t is the time it takes for the sound to travel back up. Alternatively, they can use the equation -h=-1/2*g*T^2, where T is the time it takes for the rock to fall and h is the depth of the mineshaft. By setting these equations equal to each other, the student can solve for T or t and then use that value to find the depth of the mineshaft using the equation x=0.5*a*t^2, where a=9.

## Homework Statement

A student drops a rock down a deep mine shaft, how deep is the mineshaft if he hears the rock hit the bottom 6.5s after droping it from rest?

## Homework Equations

sound in air = 343 m/s
g= 9.8 m/s

## The Attempt at a Solution

cant figure out how to get the answer.
Not sure what the correct steps are.

So you know the sound travels h=343*t, h will be the depth of the mineshaft, and t is how long it takes AFTER the rock hits the ground

the equation for the rock is -h=-1/2*g*T^2 where T is the time it takes the rock to fall down the mineshaft and h is again the depth of the mineshaft(remember that in that equation it's final position minus initial, the final is 0, the initial is h, so that's why it's negative h)

Remember that 6.5-T=t (or 6.5-t=T) so you can plug whichever into either equation and set the equations equal(h=h, watch out for the negative)and find T or t so you can find h

Last edited:
This is a free-fall problem:
the equation to be used is,

x=0.5*a*t^2
since a=g=9.8 m/s

x=0.5*9.8*t^2 = 4.9*t^2

you have t=6.5s

so x=4.9*9.8*(6.5)^2 = 207 meters.

which is how deep the mineshaft.

Free free to ask any other questions

torresmido would be correct except you're not told the rock hits 6.5 seconds later, you're told you HEAR it hit 6.5 seconds later.

So you drop it, it falls and then hits, then the sound travels back up, and that whole process takes 6.5 seconds

## 1. How do you calculate the velocity of a rock dropped down a mine shaft?

To calculate the velocity of a rock dropped down a mine shaft, we use the equation v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the mine shaft.

## 2. What is the acceleration of a rock dropped down a mine shaft?

The acceleration of a rock dropped down a mine shaft is the acceleration due to gravity, which is approximately 9.8 m/s². This means that the rock will increase in velocity by 9.8 meters per second every second it falls.

## 3. How does the mass of the rock affect its velocity when dropped down a mine shaft?

The mass of the rock does not affect its velocity when dropped down a mine shaft. According to Newton's Second Law, the acceleration of an object is directly proportional to the force applied to it, and inversely proportional to its mass. In this case, the only force acting on the rock is gravity, so the mass will not affect its acceleration or velocity.

## 4. What is the final velocity of the rock when it reaches the bottom of the mine shaft?

The final velocity of the rock when it reaches the bottom of the mine shaft can be calculated using the same equation as in question 1 (v = √(2gh)). However, in this case, the height (h) will be the depth of the mine shaft.

## 5. Is air resistance a factor when a rock is dropped down a mine shaft?

No, air resistance is not a factor when a rock is dropped down a mine shaft. This is because the rock is falling in a confined space and the air inside the mine shaft does not create enough resistance to significantly affect the rock's velocity. However, if the mine shaft is very deep, air resistance may become a factor as the rock gains more velocity.

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