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Old Sep 6th 2009, 09:23 PM   #11
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Originally Posted by Deco View Post
$\displaystyle \Delta x = V_it + \frac {1}{2}gt^2$

$\displaystyle V_i = \frac {\Delta x}{t} + \frac {1}{2}gt$
I should point out that your algebra is a little off - probably why there is the difference in answers.

$\displaystyle \Delta x = V_i t + \frac{1}{2}gt^2$

$\displaystyle V_i = \frac{\Delta x}{t} - \frac{1}{2}gt$
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Old Sep 7th 2009, 11:55 PM   #12
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I was a little surprised to find pmb's value differerent from 93.9 though the approach is nice and i couldn't spot anything wrong. So i tried to calculate using the difference in time and used the following equations.

$\displaystyle H = \frac{1}{2}gt^2$

$\displaystyle H-38 = \frac{1}{2}g(t-1)^2 + 38 $.

Solving the above i ended up with t = 4.377 secs giving me H = 93.898 as before.

I finally found that the problem was in this step



as g should multiply 1/2 also here.

With this, the answer works out to be the same i.e. 93.90

Last edited by physicsquest; Sep 8th 2009 at 12:00 AM.
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Old Sep 8th 2009, 11:25 AM   #13
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Well,

I think Eva's question is a bit misleading, she also said " at the last second, v=-38.0m/s" so should you take it that 1 sec should be added?

If so i get the answer = 116.57 m
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Old Sep 8th 2009, 12:58 PM   #14
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Well, the question stated that it fell a total of 38m in the last second, not that the velocity at the last second (or the second before last) was 38m/s. I guess you could say that, for the last second, the average velocity was -38m/s.
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Old Sep 8th 2009, 07:11 PM   #15
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I guess you could say that, for the last second, the average velocity was -38m/s.
You could because g is a constant, making it constant acceleration.
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Old Oct 9th 2009, 12:33 PM   #16
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Originally Posted by physicsquest View Post
I was a little surprised to find pmb's value differerent from 93.9 though the approach is nice and i couldn't spot anything wrong. So i tried to calculate using the difference in time and used the following equations.

$\displaystyle H = \frac{1}{2}gt^2$

$\displaystyle H-38 = \frac{1}{2}g(t-1)^2 + 38 $.

Solving the above i ended up with t = 4.377 secs giving me H = 93.898 as before.

I finally found that the problem was in this step



as g should multiply 1/2 also here.

With this, the answer works out to be the same i.e. 93.90
I did carry out the multiplication factor of 1/2 to the g term. But since there was a factor of 2 inside the parentheses multiplying $\displaystyle t_i$ it cancled the factor of two and ended up with the following is an equality.

$\displaystyle \Delta x = \frac{1}{2}g(2t_i + 1) = gt_i + \frac{1}{2}$

Do you see what I mean? This still puzzles me why we all got different answers. I've been out of sorts lately due to health issues but am starting to feel better so I'll take another look soon.
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Old Oct 9th 2009, 10:17 PM   #17
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Originally Posted by pmb
$\displaystyle
\Delta x = \frac{1}{2}g(2t_i + 1) = gt_i + \frac{1}{2}
$

It should be


$\displaystyle
\Delta\ x\ =\ gt_i\ +\ \frac{1}{2}g$

I am referring to the term 1 inside the bracket , not the $\displaystyle 2t_i$ term
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Old Oct 9th 2009, 10:49 PM   #18
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Originally Posted by physicsquest View Post
It should be


$\displaystyle
\Delta\ x\ =\ gt_i\ +\ \frac{1}{2}g$

I am referring to the term 1 inside the bracket , not the $\displaystyle 2t_i$ term
Holy gravity Batman! I can't believe what a goof I made. Man-O-man! If the problem was any closer to me it would have bit my tonsils out. Thank you my friend for pointing that out to me.
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Old Oct 9th 2009, 10:59 PM   #19
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Yeah , it got my tonsils too!
The error fits so snugly, it took me 2 days to find it!
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Old Oct 10th 2009, 02:50 AM   #20
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Physicsquest,

This is only a question and i'm probably wrong, but here goes.

If your answer of 93.898 m is correct and the final second the object travels 38m then the decimal fraction for the last sec is 0.35681 of the total displacement.

So 1 sec / (1-(sq-rt (1-0.35681)) = 5.05 secs total time falling

which would imply something is wrong, perhaps myself

Last edited by Paul46; Oct 10th 2009 at 04:29 AM.
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