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Old 09-04-2003, 02:47 PM   #1
winter
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Physics / Math question

At time = 0 a stone is dropped from the top of a cliff above a lake.

Another stone is thrown downward 1.6 seconds later from the same point with an inital speed of 32 m/s.

Both stones hit the water at the same instant. Find the height of the cliff.
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Old 09-04-2003, 02:49 PM   #2
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E=MC2

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Old 09-04-2003, 02:50 PM   #3
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Remember that the ONLY difference between the two rocks is the initial velocity. one has VI'=0 and other is VI''=32

Accelleration is the same, mass is the same (I assume), distance is the same.
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Old 09-04-2003, 02:51 PM   #4
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51.2m

(edit: first guess, but wrong)
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Old 09-04-2003, 02:52 PM   #5
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Google it. Google's supposedly got some kickass calculation software hooked into their engine.
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Old 09-04-2003, 02:52 PM   #6
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damn i don't remember the old formulas for that, it should be easy if you know them
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Old 09-04-2003, 02:52 PM   #7
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first do distance = distance
left side of equation is 1/2 g (t-1.6)^2 + 32(t-1.6)
right side is 1/2 g t^2

solve for t

then plug in that value of t to either side to find the value of height


Thats pretty close... not bad for in my head anyway.
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Old 09-04-2003, 02:53 PM   #8
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into a lake? No way to know... since a lake could be at any level above or below sea level, thus effecting the gravitational pull... altering the speed of the falling rock.

Now... had you said ocean....
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Old 09-04-2003, 02:55 PM   #9
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not enough info
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Old 09-04-2003, 02:55 PM   #10
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Is it on earth?
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Old 09-04-2003, 02:58 PM   #11
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the answer is 27.6 meteres.... I just dont know the equations to use :P
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Old 09-04-2003, 02:58 PM   #12
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Quote:
Originally posted by MaskedMan
into a lake? No way to know... since a lake could be at any level above or below sea level, thus effecting the gravitational pull... altering the speed of the falling rock.

Now... had you said ocean....
acceleration rate of gravity is 32 fps square over an ocean, lake, or dry land (not taking wind resistance into account)
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Old 09-04-2003, 03:01 PM   #13
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Quote:
Originally posted by gornyhuy
first do distance = distance
left side of equation is 1/2 g (t-1.6)^2 + 32(t-1.6)
right side is 1/2 g t^2

solve for t

then plug in that value of t to either side to find the value of height


Thats pretty close... not bad for in my head anyway.
yeah, you got it, thanks man!
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Old 09-04-2003, 03:01 PM   #14
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Quote:
Originally posted by BV


acceleration rate of gravity is 32 fps square over an ocean, lake, or dry land (not taking wind resistance into account)
Do they actually teach physics in the USA in imperial units?

That seems very very backward.
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Old 09-04-2003, 03:01 PM   #15
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84 metres?
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Old 09-04-2003, 03:03 PM   #16
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Quote:
Originally posted by winter


yeah, you got it, thanks man!
WOOO HOOOO.

Cool... Good to know that shit is still in my brain somewhere even though I never use it anymore.
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Old 09-04-2003, 03:03 PM   #17
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Quote:
Originally posted by Dusen


Do they actually teach physics in the USA in imperial units?

That seems very very backward.
No its always metric from grade school through grad school... at least it was for me.
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Old 09-04-2003, 03:03 PM   #18
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After careful consideration of all the data, I will say 3.
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Old 09-04-2003, 03:03 PM   #19
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Quote:
Originally posted by Dusen


Do they actually teach physics in the USA in imperial units?

That seems very very backward.
I don't know about now but back in 1983 they did.
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Old 09-04-2003, 03:05 PM   #20
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damn, 3 seconds late. i just forgot the formula.
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Old 09-04-2003, 03:09 PM   #21
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Touche.

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