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Old Sep 5th 2009, 08:57 PM   #1
Eva
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Unhappy Falling object

So I'm doing my physics assignment, and I totally hit a road block.
Here's the very straight forward question:
An object is dropped from height H. During the final second of its fall it traverses a distance of 38.0 m. What was H?

what I have done is the following.
vo= 0m/s
a=-9.8m/s2
at the last second, v=-38.0m/s
therefore (in my head) delta t = a* delta v
so = 372.4s

so delta x= (vo+vf)/2 * delta t
which would equal 7075.6m
which is definitely not a reasonable answer.
help? i've never been good at physics..
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Old Sep 6th 2009, 12:58 AM   #2
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Let v be the velocity of the object just before it hits the ground.

Since it was dropped (and not thrown down), we have from conservation of energy,

$\displaystyle \frac{1}{2}mv^2 = mgH$ or $\displaystyle v^2 = 2.g.H$, or

$\displaystyle v = \sqrt{2gH}$ …..(1)


Let u be the velocity it attains at the beginning of the last second.

By now it has fallen thru a height of (H -38.0) m. So by the same logic,

$\displaystyle u = \sqrt{2g(H-38)}$…… (2).

But v is the final velocity attained by the object starting with an initial velocity of u in the last 1 second. So using

$\displaystyle v= u+gt$ , we have from (1) and (2)

$\displaystyle \sqrt{2gH} = \sqrt{2g(H-38)} + g.1$ .

Solve for H with $\displaystyle g = 9.8 m/sec^2$


Spoiler:
giving H = 93.898 metres

Last edited by physicsquest; Sep 6th 2009 at 10:13 AM.
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Old Sep 6th 2009, 12:23 PM   #3
Eva
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wow.
i would've never thought of that.
thanks!
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Old Sep 6th 2009, 02:22 PM   #4
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I got something a little different, just plugging in numbers to some familiar equations:

Given:
$\displaystyle y_o=0$
$\displaystyle v_0=0$
$\displaystyle v_f=-38$
$\displaystyle a=-9.8$


$\displaystyle v_f=at+v_0$

$\displaystyle -38=-9.8t$

$\displaystyle t \approx 3.88(s)$


$\displaystyle y_f=\frac{1}{2}at^2$

$\displaystyle y_f=\frac{1}{2}(-9.8)(3.88)^2$

$\displaystyle y_f \approx -73.8(m)$


and your height H is the magnitude of that, so

$\displaystyle H \approx |-73.8|= \boxed{73.8(m)}$
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Old Sep 6th 2009, 02:28 PM   #5
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Originally Posted by nickerus View Post
I got something a little different, just plugging in numbers to some familiar equations:

Given:
$\displaystyle y_o=0$
$\displaystyle v_0=0$
$\displaystyle v_f=-38$
$\displaystyle a=-9.8$


$\displaystyle v_f=at+v_0$

$\displaystyle -38=-9.8t$

$\displaystyle t \approx 3.88(s)$


$\displaystyle y_f=\frac{1}{2}at^2$

$\displaystyle y_f=\frac{1}{2}(-9.8)(3.88)^2$

$\displaystyle y_f \approx -73.8(m)$


and your height H is the magnitude of that, so

$\displaystyle H \approx |-73.8|= \boxed{73.8(m)}$

$\displaystyle v_f=at+v_0$

v_0 does not equal 0 and we don't know v_f.
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Last edited by Deco; Sep 6th 2009 at 02:33 PM.
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Old Sep 6th 2009, 03:43 PM   #6
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Here is what I get - Let x be the distance traveled by the falling object. We know that $\displaystyle v_0 = 0$. Then

Eq. (1) $\displaystyle x = \frac{1}{2}gt^2$

where I've defined x to increase in the down direction. Let $\displaystyle t_f$ and $\displaystyle t_i$ be defined such that

$\displaystyle H = x(t_f)$

$\displaystyle t_i = t_f - 1$

or

$\displaystyle t_f = t_i + 1$

Define $\displaystyle x_f$ and $\displaystyle x_i$ as

$\displaystyle x_i = x(t_i)$

$\displaystyle x_f = x(t_f)$

From the statement of the problem we know

$\displaystyle \Delta x = x_f - x_i = 38.0 m$

From Eq. (1) we obtain

$\displaystyle x_f = \frac{1}{2}gt_f^2$
$\displaystyle x_i = \frac{1}{2}gt_i^2$

$\displaystyle \Delta x = x_f - x_i = \frac{1}{2}g({t_f^2-t_i^2})$

$\displaystyle t_f^2 = (t_i + 1)^2 = t_i^2 + 2t_i + 1$

or

$\displaystyle t_f^2 - t_i^2 = 2t_i + 1$

This gives us

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

Solve for $\displaystyle t_i$ to get

$\displaystyle t_i = \frac{1}{g}(\Delta x - \frac{1}{2}) = 3.83 s$

This gives

$\displaystyle t_f = 4.83 s$

Plug this into $\displaystyle H = x(t_f)$ to get H = 114 m.

Eva - You made an error when you said that at the last second, v=-38.0m/s. I believe you mixed up position and velocity.

Last edited by Pmb; Sep 6th 2009 at 03:48 PM.
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Old Sep 6th 2009, 03:49 PM   #7
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Right =p I should have seen that. Ok, I revised it, although I believe the initial velocity should still be 0, since it was dropped and not thrown. Anyways, here goes:

Given:
$\displaystyle a=-9.8$

So, if we imagine the ball being dropped, then we can say that it falls H meters in t seconds. We can also say that it falls (H - 38) meters in (t - 1) seconds. So, we set up a couple of equations:

$\displaystyle y(t-1)=\frac{1}{2}(-9.8)(t-1)^2$
and
$\displaystyle y(t)=\frac{1}{2}(-9.8)t^2$

we then say that
$\displaystyle y(t-1) - 38 = y(t)$

$\displaystyle (-4.9)(t-1)^2 - 38=(-4.9)t^2$
$\displaystyle (-4.9)(t^2-2t+1)=(-4.9)t^2+38$
$\displaystyle t^2-2t+1=t^2-7.755$
$\displaystyle -2t=-8.755$
$\displaystyle t \approx 4.3775$

plugging this into y(t), we get:
$\displaystyle y(4.3775)=(-4.9)(4.3775)^2 \approx -93.90$

and the magnitude is our height H, so...
$\displaystyle \boxed{H \approx 93.90m}$
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Old Sep 6th 2009, 03:57 PM   #8
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$\displaystyle \Delta x = V_it + \frac {1}{2}gt^2$

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

$\displaystyle V_i = 33.1 \mbox { m/s}$

From conservation of energy:

$\displaystyle mg(H - 38) = \frac {1}{2}mv_i^2$

$\displaystyle H - 38 = \frac {v_i^2}{2g}$

$\displaystyle H - 38 = 55.9 $

$\displaystyle H = 93.9 \mbox { m}$

Edit: Don't try to overcomplicate things by considering time of the different heights.
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Last edited by Deco; Sep 7th 2009 at 06:49 AM.
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Old Sep 6th 2009, 04:40 PM   #9
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Ok, but, if you plug your numbers into the equations and solve for t, the difference is not 1 second:

$\displaystyle 131.9=4.9t^2$
$\displaystyle t=5.18$

$\displaystyle 93.9=4.9t^2$
$\displaystyle t=4.37$

so the difference in time between those two height values is .81 seconds, unless I'm way off base here.
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Old Sep 6th 2009, 05:43 PM   #10
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Yes, but keeping in mind that that was all a rough calculation and the numbers are only off by 0.19 seconds. Remember, rough calculation.


(This is about a 3.98% error from the recorded times.)
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Last edited by Deco; Sep 6th 2009 at 06:16 PM.
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