# Find Cliff Height | Physics Homework

• Toranc3
In summary, the conversation discusses a problem where a person drops a rock from a cliff and hears the sound of it hitting the ground 10 seconds later. By using equations for the rock's fall and the sound's travel time, they are able to determine the height of the cliff to be 381.7850 meters. Part b of the problem asks whether ignoring the sound's travel time would overestimate or underestimate the height of the cliff, with the correct answer being overestimate since sound is not affected by gravity.
Toranc3

## Homework Statement

Suppose you are climbing in the high sierra when you suddenly find yourself at the edge of a fog shrouded cliff. To find the height of this cliff, you drop a rock from the top and 10 s later, hear the sound of it hitting the ground at the foot of the cliff.
a) Ignoring air resistance, how high is the cliff if the speed of sound is 330 m/s?
b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explan your reasoning.

## Homework Equations

y=yo+vo*t+1/2*a*t^(2)

-b +or- [Sqrt(b^(2)-4ac)]/2a

## The Attempt at a Solution

t1=time for the rock to fall to the ground.
t2=time for the sounds to go from bottom to top of cliff

t1+t2=10seconds

For rock falling down:

y=yo+vo*t+1/2*a*t^(2)

0=H-4.905m/s^(2)*t1^(2)

For sound coming up from the bottom:

y=yo+vo*t+1/2*a*t^(2)

H=330m/s*t2 - 4.905m/s^(2)*t2^(2)

substitute t2 with (10seconds-t1) and then equate the equations from sound and rock.

4.90m/s^(2)*t1^(2)=330m/s(10s-t1)-4.905m/s^(2)(10s-t1)^(2)

I end up with 9.81m/s^(2)*t1^(2) + 231.9m/s*t1-2809.5m

I use the quadratic formula to get this

t1=8.8224 seconds

I end up with H=381.7850m.

I do not get part b though.

Last edited:
Toranc3 said:

## Homework Statement

Suppose you are climbing in the high sierra when you suddenly find yourself at the edge of a fog shrouded cliff. To find the height of this cliff, you drop a rock from the top and 10 s later, hear the sound of it hitting the ground at the foot of the cliff.
a) Ignoring air resistance, how high is the cliff if the speed of sound is 330 m/s?
b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explan your reasoning.

## Homework Equations

y=yo+vo*t+1/*a*t^(2)

-b+- Sqrt(B^(2)-4ac)/2a
Be careful with how you write the equations. Neither of these is correct.

## The Attempt at a Solution

t1=time for the rock to fall to the ground.
t2=time for the sounds to go from bottom to top of cliff

t1=t2=10seconds
This is also not correct.
For rock falling down:

y=yo+vo*t+1/*a*t^(2)

0=H-4.905m/s^(2)*t1^(2)
When working problems, it is always best to solve the problem symbolically, and only plug in numbers at the very end. One problem with plugging in numbers in the beginning is that you force your readers to figure out what the numbers are and where they come from.
For sound coming up from the bottom:

y=yo+vo*t+1/*a*t^(2)

H=330m/s*t2 - 4.905m/s^(2)*t2^(2)
Why the second term on the right?
I do not get part b though.
What don't you get about it?

Sound is not subject to gravity!

So for sound 1/2*a*t^(2) should go to zero right? Got it thanks guys!

I would like to commend you on your attempt to solve this problem using the appropriate equations and concepts from physics. However, I would like to point out a few things that could help improve your solution.

Firstly, in your initial attempt at finding the time it takes for the rock to fall (t1), you have not taken into account the initial velocity (vo) of the rock. This should be taken into consideration as the rock is initially dropped from rest. So the correct equation would be y=yo+vo*t+1/2*a*t^(2), where yo=0 and vo=0, since the rock is dropped from rest.

Secondly, in your attempt to calculate the height of the cliff using the time for sound to reach the top of the cliff (t2), you have used the equation y=yo+vo*t+1/2*a*t^(2) incorrectly. This equation is applicable for objects in free fall, but sound does not fall, it travels at a constant speed. So the correct equation to use in this case would be y=yo+vo*t, where yo=0 and vo=330m/s.

Now, for part b of the problem, if you had ignored the time it takes for the sound to reach you, you would have underestimated the height of the cliff. This is because the time it takes for the sound to reach you adds to the total time (10 seconds) and therefore, if you only consider the time for the rock to fall, it would result in a smaller value for t1 and consequently, a smaller value for the height of the cliff (H). This can be seen in your calculation where you have used t2=10s-t1, which would result in a smaller value for t2 if t1 is underestimated.

I hope this helps clarify your doubts and provides a satisfactory response to the given content. Keep up the good work in applying physics concepts to solve real-life problems!

## What is the equation for finding the height of a cliff using physics?

The equation for finding the height of a cliff using physics is h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes for an object to fall.

## What information do I need to gather in order to use the equation to find the cliff height?

You will need to gather the time it takes for an object to fall from the top of the cliff to the bottom, as well as the acceleration due to gravity (which is a constant).

## How accurate is this method of finding the cliff height?

This method is fairly accurate, but it may not take into account factors such as air resistance or variations in the acceleration due to gravity.

## Can this equation be used for any type of cliff or only for cliffs with a straight drop?

This equation can only be used for cliffs with a straight drop. It does not take into account cliffs with varying angles or surfaces.

## What other methods can be used to find the height of a cliff?

Other methods include using trigonometry and measuring the angle of elevation from a known distance, or using more advanced technologies such as LiDAR or satellite imagery.

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