If a rock is dropped off a sea cliff

kbrowne29

If a rock is dropped off a sea cliff, and the sound of it striking the water is heard 3.4 seconds later, how high is the cliff, assuming the speed of sound to be 340 m/s.
I don't really know where to begin with this one, and I would really appreciate the help. Thanks.

mathman

3 equations in 3 unknowns. You do the algebra.

1) h=gs²/2
2) h=vt
3) s+t=3.4

Symbols:
h=height of cliff
g=gravitational constant (9.83 m/s²)
s=time for rock to fall
t=time for sound to come back to you
v=speed of sound (340 m/s)

(h,s,t) are unknowns.

kbrowne29

Thanks for the help. As I said before, my difficulty lay in not being able to start the problem, and now I know where I must begin. Thanks again for helping me out.

lavalamp

It's intersting that your value for g is 9.83. Was that a typo or is that what you've been given. I've always been told that it was 9.81.

Pandemonium

You live in England, so the gravitational acceleration there is 9.81 m s^-2. In other places the value of g is different, due to the fact that Earth is not a perfect sphere.

Where I live, in Singapore which is near the equator, the value of g is 9.78 m s^-2, although in classrooms we use 9.81 m s^-2 because the public education here prepares us for Cambridge papers.

lavalamp

I know that the earth is not a perfect sphere, and that invariably gravity will vary. But I thought that 9.81 was an average figure denoted by SI. After all, they set the standard units for everything else so why not a standard value for g aswell.

Pandemonium

Hmmm... I've never thought of that. I don't know...

Adrian Baker

Originally posted by lavalamp
I know that the earth is not a perfect sphere, and that invariably gravity will vary. But I thought that 9.81 was an average figure denoted by SI. After all, they set the standard units for everything else so why not a standard value for g aswell.

Because the other things that they set values for are constants (or are believed to be), whereas 'g' certainly isn't. If I measure 'g'in the lab, I want to find the value of it, not see how close I am to some arbitrary average decided on by a committee.
[:)]

HallsofIvy

g is not an arbitrary unit like "1 kg" or "1 meter". It is a "constant of nature" and the correct value, to the necessary accuracy, should be used for the position on the earth. Of course, for most applications, 9.81 gives the necessary accuracy. (And, in fact, 9.8 works nicely most of the time.)

russ_watters

I hadn't realized it changed so much so I looked it up. The difference is a combination of distance from the center of the earth and centrifugal force. I don't feel like doing the math right now on exactly how much of it is due to each.

lavalamp

The formulae is:

F' = GMm
---
r^2

F' is the force of attraction between two point masses of M and m.
G is the gravitational constant (6.67*10^24).
r is the distance between the two point masses.

You could take account of the fact that we're trying to mave in a straihg line and the earth is pulling us round in a circle if you wanted to, but this effect would be smal when you take account of the radius of the earth: 40,000,000/(2[pi]), nevertheless the formula for that is:

F'' = mr * ω^2

where &omega is the angular velocity, which can be replaced by 2[pi]/T where T is the time persiod 24*3600 (24 hours in seconds), m is your mass by the way:

F'' = mr * 4 * [pi]^2 / (24*3600)^2

So the total resultant force on a mass m, F = F' - F''

Feel free to work out the values if you can be bothered.

Julian Solos

Originally posted by HallsofIvy
g is not an arbitrary unit like "1 kg" or "1 meter". It is a "constant of nature" and the correct value, to the necessary accuracy, should be used for the position on the earth.

Excuse me, but g is not a constant. It is commonly called the acceleration due to gravity, which is related to the Gravitational constant, G. The value of g depends on where it is measured, including the altitude.

Even though g is commonly called the acceleration due to gravity, the measured (and published) value of g at a certain location usually includes the effect due to the rotation of the Earth at the particular location.

PrudensOptimus

the height is 578m.

Δy = [(v+v₀)/2]t

v = 0, t = 3.4s.

Magg$

Its Magg$,

you know, we talked before and you helped me with my coursework...

Only this time I'm talking out of my own interest...

Could you tell me what the Young's Modulus value is for:
Graphite,
Wood,
Steel,
Glass,
Aluminium,
and Titanium...

I know your gonna think this is to help with coursework again because Young's Modulus is part of the AS syllabus but HONESTLY, this is through my own interest as I want more young's modulus values to do some comparisons between materials.

Please Help,

Magg$

P.S, I'm going to a lecture tomorrow about the Physics of Skydiving at Birmingham University, should be good!

selfAdjoint

Here's a site with some of them. You can probably find more with a little judicious googling. I found this one just by googling on Young's Modulus. You should generally try that kind of thing before asking here, because you get the answer quicker and (hem) learn to do independent research.

Similar Threads for: If a rock is dropped off a sea cliff
Thread	Forum	Replies
Error measurement in time of flight tests	Classical Physics	4
Freefall	Introductory Physics Homework	2
Freefall	General Physics	6
Freefall	Introductory Physics Homework	3
Freefall	Introductory Physics Homework	3

kbrowne29	If a rock is dropped off a sea cliff Tweet If a rock is dropped off a sea cliff, and the sound of it striking the water is heard 3.4 seconds later, how high is the cliff, assuming the speed of sound to be 340 m/s. I don't really know where to begin with this one, and I would really appreciate the help. Thanks.

mathman Recognitions: Science Advisor	3 equations in 3 unknowns. You do the algebra. 1) h=gs²/2 2) h=vt 3) s+t=3.4 Symbols: h=height of cliff g=gravitational constant (9.83 m/s²) s=time for rock to fall t=time for sound to come back to you v=speed of sound (340 m/s) (h,s,t) are unknowns.

lavalamp	If a rock is dropped off a sea cliff It's intersting that your value for g is 9.83. Was that a typo or is that what you've been given. I've always been told that it was 9.81.

Pandemonium	You live in England, so the gravitational acceleration there is 9.81 m s^-2. In other places the value of g is different, due to the fact that Earth is not a perfect sphere. Where I live, in Singapore which is near the equator, the value of g is 9.78 m s^-2, although in classrooms we use 9.81 m s^-2 because the public education here prepares us for Cambridge papers.

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	g is not an arbitrary unit like "1 kg" or "1 meter". It is a "constant of nature" and the correct value, to the necessary accuracy, should be used for the position on the earth. Of course, for most applications, 9.81 gives the necessary accuracy. (And, in fact, 9.8 works nicely most of the time.)

russ_watters Mentor	I hadn't realized it changed so much so I looked it up. The difference is a combination of distance from the center of the earth and centrifugal force. I don't feel like doing the math right now on exactly how much of it is due to each.

PrudensOptimus	the height is 578m. Δy = [(v+v₀)/2]t v = 0, t = 3.4s.

selfAdjoint Recognitions: Gold Member Staff Emeritus	Here's a site with some of them. You can probably find more with a little judicious googling. I found this one just by googling on Young's Modulus. You should generally try that kind of thing before asking here, because you get the answer quicker and (hem) learn to do independent research.