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anonymous
 5 years ago
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
 5 years ago
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?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now distance travelled in the last second (t) is h/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02h/3= g[(t^2  (t1)^2)}

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not realy sure where you are getting the {t^2(t1)^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
 5 years ago
Best ResponseYou've already chosen the best response.0from the first eqn get t, which is is the total time traveled

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then the distance in the last second, is the distance in t seconds  distance in t1 seconds right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so h/3 = [gt^2  g(t1)^2] / 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so now uve got 2 eqns and two variables t and h

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0eliminate t from the system of equations and get h

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is around 173 meters

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and him, your method is incorrect, btw.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02h/3= g[(t^2  (t1)^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
 5 years ago
Best ResponseYou've already chosen the best response.0tht gets subtracted neway

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0linalg009, is that the correct answer in your book?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0learn to read..i explained it above

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0oops....I mean get the right distance for the last second of the fall ha....i ration of 1/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0wtdu hav to find out temme clearly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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
 5 years ago
Best ResponseYou've already chosen the best response.0ive told u the method..wat part do u not get? im positive its correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What are the two equations? 2h=gt^2 and 2h/3= g[(t^2  (t1)^2)}..?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or first u divide the two to make a quadratic eqn in t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then get the value of t and plug it into the first eqn to get h..got it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, him is right. I did not read what he had posted correctly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is around 145150 meters.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Perfect! Thanks for the help him1618 and dhatraditya ha..it makes perfect sense now! ha.
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