anonymous
  • anonymous
A rock is dropped from a cliff. How high is the cliff if the rock falls (1/3) of the total height of the cliff in the last second of it's fall?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
2h=gt^2
anonymous
  • anonymous
now distance travelled in the last second (t) is h/3
anonymous
  • anonymous
2h/3= g[(t^2 - (t-1)^2)}

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anonymous
  • anonymous
that should do it
anonymous
  • anonymous
got it?
anonymous
  • anonymous
I'm not realy sure where you are getting the {t^2-(t-1)^2} to be honest. I understand 2h=gt^2.....but then it looks like you divided that by 3 and then didn't do the same to the otherside ha......wait...maybe I do see what you did..you constrained t so that however long it takes for rock to fall 2/3 the height, it's always 1 less than the total fall time....but i'm still not sure how this will give me the cliff's height ha
anonymous
  • anonymous
from the first eqn get t, which is is the total time traveled
anonymous
  • anonymous
then the distance in the last second, is the distance in t seconds - distance in t-1 seconds right?
anonymous
  • anonymous
so h/3 = [gt^2 - g(t-1)^2] / 2
anonymous
  • anonymous
so now uve got 2 eqns and two variables t and h
anonymous
  • anonymous
eliminate t from the system of equations and get h
anonymous
  • anonymous
got it?
anonymous
  • anonymous
the answer is around 173 meters
anonymous
  • anonymous
and him, your method is incorrect, btw.
anonymous
  • anonymous
how com?
anonymous
  • anonymous
2h/3= g[(t^2 - (t-1)^2)} how did you get this? also, in the last second of its fall, the rock already has an initial velocity. You are discounting that.
anonymous
  • anonymous
tht gets subtracted neway
anonymous
  • anonymous
linalg009, is that the correct answer in your book?
anonymous
  • anonymous
learn to read..i explained it above
anonymous
  • anonymous
I almost feel like I'd have to trial and error it until I get around the correct time for the last second of the fall. ha....It doesn't have an answer in the book :/
anonymous
  • anonymous
oops....I mean get the right distance for the last second of the fall ha....i ration of 1/3
anonymous
  • anonymous
ratio*
anonymous
  • anonymous
so now the qustn is?
anonymous
  • anonymous
The question is...can I just plug in different values for the initial height of the drop until I get around 1 sec. for the last third of the fall ha..somehow
anonymous
  • anonymous
wtdu hav to find out temme clearly
anonymous
  • anonymous
It's the same question. I'm trying to find out the height of the cliff. I was just trying to think of a different method
anonymous
  • anonymous
ive told u the method..wat part do u not get? im positive its correct
anonymous
  • anonymous
What are the two equations? 2h=gt^2 and 2h/3= g[(t^2 - (t-1)^2)}..?
anonymous
  • anonymous
yeah right
anonymous
  • anonymous
t=sqrt(2h/g)
anonymous
  • anonymous
or first u divide the two to make a quadratic eqn in t
anonymous
  • anonymous
then get the value of t and plug it into the first eqn to get h..got it?
anonymous
  • anonymous
ya...solving now ha
anonymous
  • anonymous
yes, him is right. I did not read what he had posted correctly.
anonymous
  • anonymous
the answer is around 145-150 meters.
anonymous
  • anonymous
I get 145.52 meters
anonymous
  • anonymous
yes, thats right.
anonymous
  • anonymous
Perfect! Thanks for the help him1618 and dhatraditya ha..it makes perfect sense now! ha.

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