How high is the cliff and speed of sound?

In summary, the problem involves finding the height of a cliff by dropping a rock from the top and measuring the time it takes for the sound of the rock hitting the ground to travel back up to the top of the cliff. Using equations for the rock's motion and the sound's motion, a system of equations can be set up to solve for the height of the cliff. By setting the equations equal to each other and simplifying, the time it takes for the rock to fall can be determined and used to solve for the height of the cliff.
  • #1
katamoria
15
0

Homework Statement


To find the height of a cliff, you drop a rock from the top, and 10s later, hear the sound of it hitting the ground at the foot of the cliff.
ignoring air resistance, how high is the cliff if the speed of sound is 330m/s



Homework Equations



Xf = Xo +Vo(t) + 1/2at^2
Vf = Vo + at
Vf^2 = Vo^2 + 2a(Xf - Xo)

The Attempt at a Solution


I don't know how to account for the time it takes sound to travel. If I solve for Vf or Xf using these equations, then I get a number I know is for the ball being lower than the foot of the cliff. When I tried, using speed of sound, I don't know how long the sound is travelling, or how far it has to go.
 
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  • #2
Do you have an equation relating the speed of sound, the height of the cliff, and how long it takes the sound to travel the height of the cliff?
 
  • #3
Since it is all in one direction, straight path, the speed of sound is the velocity. If I plug that in, I still have two variables, time it takes for sound to travel back up the cliff, and how high the cliff is.

it is 10s total for ball to hit bottom, and sound to come back up, I can't just plug in 10s.
 
  • #4
katamoria said:
Since it is all in one direction, straight path, the speed of sound is the velocity. If I plug that in, I still have two variables, time it takes for sound to travel back up the cliff, and how high the cliff is.

it is 10s total for ball to hit bottom, and sound to come back up, I can't just plug in 10s.

You're correct that there are two variables, the ones you said.

However, there are two equations you can write. One is for the rock dropping from the top of the cliff with an acceleration of -g. The other equation is for the sound traveling up the cliff at a constant speed of 330m/s. Writing out those two equations, and working with them, is how to solve this problem.
 
  • #5
ok so here is what i tried.
i drew it so that
T1 = time for rock to fall
T2 = time for sound to go back
Tt = Total time = 10s
so
T1+T2 = Tt = 10s

then I did
Vs = velocity of sound, so
Vs = deltaX / deltaT
solved for deltaX which = height of the cliff and got

deltaX = Vs(Tt-T1)

then for the rock,
Vf^2 = Vo^2 +2a(deltaX)
knowing that Vo^2 = 0, and a = g, and solving for deltaX
deltaX = Vf^2/2g

since both equations are solving for the same deltaX, I set them equal to each other
Vs(Tt-T1) = Vf^2/2g

the unknowns are T1, and Vf

I feel pretty good up to this point, then I'm not so sure, but next i tried...
to solve for T2 by
Vf = Vo +at
t is for the rock so...
Vf = g(Tt-T2)

then plug that in and get the equation

Vs(Tt-T1) = g(Tt-T2)^2/2g
but there's still two unknowns with T1 and T2, so...
if (Tt-T2) = T1, can i plug that in on the right side, so that T1 is the only variable? it's just on both sides?
 
  • #6
I really don't think i did that right...
 
  • #7
You have made a lot of progress and are nearly there (I think) :smile:

katamoria said:
ok so here is what i tried.
i drew it so that
T1 = time for rock to fall
T2 = time for sound to go back
Tt = Total time = 10s
so
T1+T2 = Tt = 10s

then I did
Vs = velocity of sound, so
Vs = deltaX / deltaT
solved for deltaX which = height of the cliff and got

deltaX = Vs(Tt-T1)

then for the rock,
Vf^2 = Vo^2 +2a(deltaX)
knowing that Vo^2 = 0, and a = g, and solving for deltaX
deltaX = Vf^2/2g

since both equations are solving for the same deltaX, I set them equal to each other
Vs(Tt-T1) = Vf^2/2g

the unknowns are T1, and Vf

I feel pretty good up to this point, then I'm not so sure, but next i tried...
to solve for T2 by
Vf = Vo +at
t is for the rock so...
Vf = g(Tt-T2)
Okay, or you could also say

Vf = g T1​

then plug that in and get the equation

Vs(Tt-T1) = g(Tt-T2)^2/2g
Small error: the right hand side here should be

g^2 (Tt-T2)^2 / 2g​

since you are really substituting

Vf^2 → g^2 (Tt-T2)^2​

but there's still two unknowns with T1 and T2, so...
if (Tt-T2) = T1, can i plug that in on the right side, so that T1 is the only variable? it's just on both sides?

Yes, you can definitely do that ... and you should get the same thing as if you had used

Vf = g T1​

earlier.

Looks like you are nearly done, just need to solve for T1.
 

1. How do you measure the height of a cliff?

The height of a cliff can be measured using various methods such as trigonometry, GPS, or laser rangefinders. A common method is to use a clinometer, which measures the angle of elevation from the base of the cliff to the top. Using this angle and the known distance from the observer to the base of the cliff, the height of the cliff can be calculated using simple trigonometry.

2. Why is the speed of sound affected by the height of a cliff?

The speed of sound is affected by the height of a cliff because sound waves travel through air molecules. As the altitude increases, the air becomes less dense, which results in a decrease in the speed of sound. This is because there are fewer air molecules for the sound waves to travel through, causing them to move slower.

3. How does temperature affect the speed of sound?

Temperature affects the speed of sound because it affects the density of air. As temperature increases, air molecules move faster and spread out, resulting in a decrease in the density of air. This decrease in density causes sound waves to travel slower. On the other hand, as temperature decreases, air molecules slow down and become more tightly packed, resulting in an increase in the speed of sound.

4. What is the formula for calculating the speed of sound?

The formula for calculating the speed of sound is: speed of sound (m/s) = 331.4 + 0.6 x temperature (°C). This formula is based on the speed of sound at sea level and standard temperature (15°C). However, this formula is just an approximation and the speed of sound can vary depending on factors such as humidity and altitude.

5. How does the speed of sound compare to the speed of light?

The speed of sound is significantly slower than the speed of light. Sound travels at approximately 343 meters per second, while light travels at a speed of approximately 299,792,458 meters per second. This means that light can travel around the Earth 7.5 times in one second, while sound can only travel a few hundred meters in the same amount of time.

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