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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?

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  1. anonymous
    • 5 years ago
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    2h=gt^2

  2. anonymous
    • 5 years ago
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    now distance travelled in the last second (t) is h/3

  3. anonymous
    • 5 years ago
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    2h/3= g[(t^2 - (t-1)^2)}

  4. anonymous
    • 5 years ago
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    that should do it

  5. anonymous
    • 5 years ago
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    got it?

  6. anonymous
    • 5 years ago
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    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

  7. anonymous
    • 5 years ago
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    from the first eqn get t, which is is the total time traveled

  8. anonymous
    • 5 years ago
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    then the distance in the last second, is the distance in t seconds - distance in t-1 seconds right?

  9. anonymous
    • 5 years ago
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    so h/3 = [gt^2 - g(t-1)^2] / 2

  10. anonymous
    • 5 years ago
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    so now uve got 2 eqns and two variables t and h

  11. anonymous
    • 5 years ago
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    eliminate t from the system of equations and get h

  12. anonymous
    • 5 years ago
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    got it?

  13. anonymous
    • 5 years ago
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    the answer is around 173 meters

  14. anonymous
    • 5 years ago
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    and him, your method is incorrect, btw.

  15. anonymous
    • 5 years ago
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    how com?

  16. anonymous
    • 5 years ago
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    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.

  17. anonymous
    • 5 years ago
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    tht gets subtracted neway

  18. anonymous
    • 5 years ago
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    linalg009, is that the correct answer in your book?

  19. anonymous
    • 5 years ago
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    learn to read..i explained it above

  20. anonymous
    • 5 years ago
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    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 :/

  21. anonymous
    • 5 years ago
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    oops....I mean get the right distance for the last second of the fall ha....i ration of 1/3

  22. anonymous
    • 5 years ago
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    ratio*

  23. anonymous
    • 5 years ago
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    so now the qustn is?

  24. anonymous
    • 5 years ago
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    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

  25. anonymous
    • 5 years ago
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    wtdu hav to find out temme clearly

  26. anonymous
    • 5 years ago
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    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

  27. anonymous
    • 5 years ago
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    ive told u the method..wat part do u not get? im positive its correct

  28. anonymous
    • 5 years ago
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    What are the two equations? 2h=gt^2 and 2h/3= g[(t^2 - (t-1)^2)}..?

  29. anonymous
    • 5 years ago
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    yeah right

  30. anonymous
    • 5 years ago
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    t=sqrt(2h/g)

  31. anonymous
    • 5 years ago
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    or first u divide the two to make a quadratic eqn in t

  32. anonymous
    • 5 years ago
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    then get the value of t and plug it into the first eqn to get h..got it?

  33. anonymous
    • 5 years ago
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    ya...solving now ha

  34. anonymous
    • 5 years ago
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    yes, him is right. I did not read what he had posted correctly.

  35. anonymous
    • 5 years ago
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    the answer is around 145-150 meters.

  36. anonymous
    • 5 years ago
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    I get 145.52 meters

  37. anonymous
    • 5 years ago
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    yes, thats right.

  38. anonymous
    • 5 years ago
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    Perfect! Thanks for the help him1618 and dhatraditya ha..it makes perfect sense now! ha.

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