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Math Help - Free Fall Lmit/Derivative problem

  1. #1
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    Free Fall Lmit/Derivative problem

    An object is dropped from the top of a 100-m tower. Its height above the ground after 1 sec is 100-4.9t^2 m. How fast is it falling 2 sec after it is dropped?

    Formula:
    lim as h--> 0 (f(t+h)- f(t))/h
    or as an easier formula, I think.
    lim as h--> 0(f(t) - f(h))/(t-h)

    Guess: 100=4.9t^2 m goes as t and solve. Then plug in 2 for t.

    Thank you!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    An object is dropped from the top of a 100-m tower. Its height above the ground after 1 sec is 100-4.9t^2 m. How fast is it falling 2 sec after it is dropped?

    Formula:
    lim as h--> 0 (f(t+h)- f(t))/h
    or as an easier formula, I think.
    lim as h--> 0(f(t) - f(h))/(t-h)

    Guess: 100=4.9t^2 m goes as t and solve. Then plug in 2 for t.

    Thank you!
    the problem says the formula is for the height after 1 seconds. doesn't that mean we plug in t = 1 for 2 seconds, since it is one more second after the formula applies? i don't know, that's oddly worded. anyway, what you want is:

    \lim_{h \to 0} \frac {f(1 + h) - f(1)}{h}

    i used 1 here for 1 additional second. maybe you should use 2 instead.
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  3. #3
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    Ah. Syntaz LaTex error.
    *you are seeing Jhevon's formula quoted*

    How do I get rid of the h on the bottom? My book peruses by it without telling.
    Something about 1/h times the whole thing.
    The answer in the book is: 19.6 m/sec.
    After I figure out the way above equation, do I put 100-4.9t^2 as a and solve?
    Thanks.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    Ah. Syntaz LaTex error.
    *you are seeing Jhevon's formula quoted*

    How do I get rid of the h on the bottom? My book peruses by it without telling.
    Something about 1/h times the whole thing.
    The answer in the book is: 19.6 m/sec.
    After I figure out the way above equation, do I put 100-4.9t^2 as a and solve?
    Thanks.
    ok. so we should plug in t = 2 then. fine.

    (did you try to work it out? it seems pretty straight forward to me)

    \lim_{h \to 0} \frac {f(2 + h) - f(2)}h

    = \lim_{h \to 0} \frac {100 - 4.9(2 + h)^2 - 80.4}h

    = \lim_{h \to 0} \frac {19.6 - 4.9 \left(4 + 4h + h^2 \right)}h

    = \lim_{h \to 0} \frac {-19.6h - 4.9h^2}h

    = \lim_{h \to 0} (-19.6 - 4.9h)

    = -19.6

    (the velocity is negative because we are falling, that is, moving in the negative direction)
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  5. #5
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    I got it now.
    It was all because in - f(a), why I'm being clueless and asking odd question.
    I would write the equation as - 100 -4.9(2)^2.
    Instead of, -(100-4.9(2)^2.

    there we go. The answer to half the problems!

    Thanks you. Yes, thanks you.
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  6. #6
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    I spoke too soon.

    You lost me here.





    how did you do that?
    It seems you stuck 4h with h as 0 to cancel it out. After that, I don't know what you did.
    Thank you.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post


    I spoke too soon.

    You lost me here.





    how did you do that?
    It seems you stuck 4h with h as 0 to cancel it out. After that, I don't know what you did.
    Thank you.
    fine, i'll show all the steps. i assume you understand everything up to that point.

    \lim_{h \to 0} \frac {19.6 - 4.9 \left( 4 + 4h + h^2 \right)}h ..........expand the brackets

    = \lim_{h \to 0} \frac {19.6 - 19.6 - 19.6h - 4.9h^2}h ........simplify

    = \lim_{h \to 0} \frac {-19.6h - 4.9h^2}h ...........factor out the h

    = \lim_{h \to 0} \frac {h(-19.6 - 4.9h)}h ............cancel the h

    = \lim_{h \to 0} (-19.6 - 4.9h) ..............take the limit

    = -19.6
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