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?

- anonymous

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- anonymous

2h=gt^2

- anonymous

now distance travelled in the last second (t) is h/3

- anonymous

2h/3= g[(t^2 - (t-1)^2)}

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## More answers

- anonymous

that should do it

- anonymous

got it?

- 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

from the first eqn get t, which is is the total time traveled

- anonymous

then the distance in the last second, is the distance in t seconds - distance in t-1 seconds right?

- anonymous

so h/3 = [gt^2 - g(t-1)^2] / 2

- anonymous

so now uve got 2 eqns and two variables t and h

- anonymous

eliminate t from the system of equations and get h

- anonymous

got it?

- anonymous

the answer is around 173 meters

- anonymous

and him, your method is incorrect, btw.

- anonymous

how com?

- 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

tht gets subtracted neway

- anonymous

linalg009, is that the correct answer in your book?

- anonymous

learn to read..i explained it above

- 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

oops....I mean get the right distance for the last second of the fall ha....i ration of 1/3

- anonymous

ratio*

- anonymous

so now the qustn is?

- 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

wtdu hav to find out temme clearly

- 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

ive told u the method..wat part do u not get? im positive its correct

- anonymous

What are the two equations? 2h=gt^2 and 2h/3= g[(t^2 - (t-1)^2)}..?

- anonymous

yeah right

- anonymous

t=sqrt(2h/g)

- anonymous

or first u divide the two to make a quadratic eqn in t

- anonymous

then get the value of t and plug it into the first eqn to get h..got it?

- anonymous

ya...solving now ha

- anonymous

yes, him is right. I did not read what he had posted correctly.

- anonymous

the answer is around 145-150 meters.

- anonymous

I get 145.52 meters

- anonymous

yes, thats right.

- anonymous

Perfect! Thanks for the help him1618 and dhatraditya ha..it makes perfect sense now! ha.

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