an unnaturally high cliff: find height, given rate of travel and speed of sound

A

allegansveritatem

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Here is the problem:


72 Height of a cliff
When a rock is dropped from a cliff into an ocean, it travels approximately 16t2 feet in t seconds. If the splash is heard 4 seconds later and the speed of sound is 1100 ft/sec, approximate the height of the cliff.


I treated this as a variant of T=D/R problem and here is what I did:

11669

This result doesn't seem possible but I'm not subtle-minded enough to come up with anything else.
 

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Google "When a rock is dropped from a cliff into an ocean, it travels approximately 16t^2 feet in t seconds"
 
I would let \(t_1\) be the time the rock fell (in seconds), and so the height \(h\) of the cliff (in feet) is:

[MATH]h=16t_1^2[/MATH]
I would let \(t_2\) be the time the sound of the splash was traveling and write:

[MATH]h=1100t_2[/MATH]
Now, we know:

[MATH]t_1+t_2=4\implies t_2=4-t_1[/MATH]
And so we have the two equations:

[MATH]h=16t_1^2[/MATH]
[MATH]h=1100(4-t_1)=4400-1100t_1\implies t_1=\frac{4400-h}{1100}[/MATH]
Substituting for \(t_1\) into the first equation, we get the quadratic in \(h\):

[MATH]h=16\left(\frac{4400-h}{1100}\right)^2[/MATH]
Now, we know:

[MATH]t_1<4\implies h<256[/MATH]
And so the only root of the above quadratic we can use is:

[MATH]h=\frac{140800}{307+5\sqrt{3729}}\approx229.942237876930[/MATH]
 
MarkFL said:
I would let \(t_1\) be the time the rock fell (in seconds), and so the height \(h\) of the cliff (in feet) is:

[MATH]h=16t_1^2[/MATH]
I would let \(t_2\) be the time the sound of the splash was traveling and write:

[MATH]h=1100t_2[/MATH]
Now, we know:

[MATH]t_1+t_2=4\implies t_2=4-t_1[/MATH]
And so we have the two equations:

[MATH]h=16t_1^2[/MATH]
[MATH]h=1100(4-t_1)=4400-1100t_1\implies t_1=\frac{4400-h}{1100}[/MATH]
Substituting for \(t_1\) into the first equation, we get the quadratic in \(h\):

[MATH]h=16\left(\frac{4400-h}{1100}\right)^2[/MATH]
Now, we know:

[MATH]t_1<4\implies h<256[/MATH]
And so the only root of the above quadratic we can use is:

[MATH]h=\frac{140800}{307+5\sqrt{3729}}\approx229.942237876930[/MATH]
Click to expand...
thanks for working through this. I will have to go over it more minutely tomorrow. For me, it is too much to take in casually given that it is an approach I didn't even know was on offer. I must say, in the context of the other parameters of the situation, 256 has a lot more going for it than 4000 and something. I mean, how does a dropped rock and a sound travel almost a mile in 4 seconds? The rock would have to have been shot from a cannon. haha I will come back when I have worked this out again and present photographic evidence of my results.
 
Just to be different (and to use the distance as the variable, as in the wrong solution), I solved [MATH]x=16t^2[/MATH] for t, [MATH]t = \frac{\sqrt{x}}{4}[/MATH], and wrote an equation for the sum of the two times, [MATH]\frac{x}{1100} + \frac{\sqrt{x}}{4} = 4[/MATH]. It's not hard to solve this radical equation, and I get the same result, 229.9 ft.

Note that the second term is very different, because it is not t = d/r at all! There is no fixed speed.
 
Dr.Peterson said:
Just to be different (and to use the distance as the variable, as in the wrong solution), I solved [MATH]x=16t^2[/MATH] for t, [MATH]t = \frac{\sqrt{x}}{4}[/MATH], and wrote an equation for the sum of the two times, [MATH]\frac{x}{1100} + \frac{\sqrt{x}}{4} = 4[/MATH]. It's not hard to solve this radical equation, and I get the same result, 229.9 ft.

Note that the second term is very different, because it is not t = d/r at all! There is no fixed speed.
Click to expand...
Thanks for posting this. I will try it tomorrow. I had a hard time with the solution posted by Mark...which I will document in the following post.
 
MarkFL said:
I would let \(t_1\) be the time the rock fell (in seconds), and so the height \(h\) of the cliff (in feet) is:

[MATH]h=16t_1^2[/MATH]
I would let \(t_2\) be the time the sound of the splash was traveling and write:

[MATH]h=1100t_2[/MATH]
Now, we know:

[MATH]t_1+t_2=4\implies t_2=4-t_1[/MATH]
And so we have the two equations:

[MATH]h=16t_1^2[/MATH]
[MATH]h=1100(4-t_1)=4400-1100t_1\implies t_1=\frac{4400-h}{1100}[/MATH]
Substituting for \(t_1\) into the first equation, we get the quadratic in \(h\):

[MATH]h=16\left(\frac{4400-h}{1100}\right)^2[/MATH]
Now, we know:

[MATH]t_1<4\implies h<256[/MATH]
And so the only root of the above quadratic we can use is:

[MATH]h=\frac{140800}{307+5\sqrt{3729}}\approx229.942237876930[/MATH]
Click to expand...
I went at it long and strong today trying to works this out. I was able to follow you this far:
11678

I was OK until I tried to solve the h=16(4400-h/1100)^2. l tried and tried to make this come to something reasonable but l always wound up in that far away country where the men have two heads and ride on dragons. As, for instance here:
11679

and here:11680

So...how do you solve 16(4400-h/1100)^2?
 
allegansveritatem said:
So...how do you solve 16(4400-h/1100)^2?
Click to expand...
That should be:
how do you solve h = 16((4400 - h) / 1100)^2
Try again; start this way:
h / 16 = ((4400 - h) / 1100)^2

I'd let u = 4400 and v = 1100; so equation becomes:
h / 16 = ((u - h) / v)^2

Then substitute back in at end...save you lots of time...

You should end up with: h^2 - 84425h + 19360000 = 0

NOTE: your handwriting is worse than mine!!
 
We could also begin with:

[MATH]h=16\left(\frac{4400-h}{1100}\right)^2[/MATH]
This implies:

[MATH]\frac{4400-h}{1100}=4-\frac{16}{1100}\left(\frac{4400-h}{1100}\right)^2[/MATH]
Let [MATH]u=\frac{4400-h}{1100}[/MATH] and arrange as:

[MATH]\frac{16}{1100}u^2+u-4=0[/MATH]
Discard the negative root, and we have:

[MATH]u=\frac{4400-h}{1100}=\frac{5\sqrt{3729}-275}{8}[/MATH]
Solve for \(h\):

[MATH]h=4400-\frac{1100}{8}(5\sqrt{3729}-275)=\frac{275}{2}(307-5\sqrt{3729})\approx229.94223787693[/MATH]
 
MarkFL said:
We could also begin with:

[MATH]h=16\left(\frac{4400-h}{1100}\right)^2[/MATH]
This implies:

[MATH]\frac{4400-h}{1100}=4-\frac{16}{1100}\left(\frac{4400-h}{1100}\right)^2[/MATH]
Let [MATH]u=\frac{4400-h}{1100}[/MATH] and arrange as:

[MATH]\frac{16}{1100}u^2+u-4=0[/MATH]
Discard the negative root, and we have:

[MATH]u=\frac{4400-h}{1100}=\frac{5\sqrt{3729}-275}{8}[/MATH]
Solve for \(h\):

[MATH]h=4400-\frac{1100}{8}(5\sqrt{3729}-275)=\frac{275}{2}(307-5\sqrt{3729})\approx229.94223787693[/MATH]
Click to expand...
It took me a little to how you derived the implication but I think I see it. I will give this a shot tomorrow. Thanks for clarifying what you meant by "quadratic equation" in your first post.
 
Denis said:
That should be:
how do you solve h = 16((4400 - h) / 1100)^2
Try again; start this way:
h / 16 = ((4400 - h) / 1100)^2

I'd let u = 4400 and v = 1100; so equation becomes:
h / 16 = ((u - h) / v)^2

Then substitute back in at end...save you lots of time...

You should end up with: h^2 - 84425h + 19360000 = 0

NOTE: your handwriting is worse than mine!!
Click to expand...
I didn't even consider the substitution method, mainly, I suppose, because I haven't been much exposed to it. I've done a few simple problems with it but...I have to do dozens of problems with anything new to get the hang of it at all. Thanks for the tip.
 
Nothing to prevent you from quickly ckecking your quadratic equation
results by using an online calculator site, like:

Quadratic Equation Calculator

Free math lessons and math homework help from basic math to algebra, geometry and beyond. Students, teachers, parents, and everyone can find solutions to their math problems instantly.
www.math.com
Enter A=1, B=-84425, C=19360000 to get:
~229.94223 or ~84195.05776
Only reason I'm mentioning both solutions:
84195/230 = 366 (a leap year!)
Perhaps Mark could explain that:)
The pebble goes around the earth and comes back in 366 days?
 
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I went back at it today and tried all the methods suggested. I kept running into problems. I tried each way multiple times and kept getting different results. Finally it occurred to me that I was so lost because I had lost sight of what I was looking for. I would get one result and have to ask myself, now is this t or is this h? Anyway, I, in extreme despair, at last went back to the original problem and read it through a few times. Then I drew a diagram. From said diagram I saw that there were two t's that when multiplied by different quantities could be made to equal h. Then I recalled from Mark's post that the t's could be expressed one in terms of the other and...the rest is presented below. I am grateful to everyone who contributed to this thread. So, after more than 5 hours over three days of head butting, I think I have broken through thus:

11688
 
I'm glad you stuck with it, and triumphed! :)

To succeed in math, you've definitely got to be tenacious.
 
HOLY!!
Suggestion: try to simplify as much as possible.

At this point in your calculations:
16(4 - t)^2 = 1100t
you could have divided by 4 to get:
4(4 - t)^2 = 275t
 
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