Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 
Sep 5th 2009, 08:57 PM

#1  Junior Member
Join Date: Sep 2009
Posts: 2
 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..

 
Sep 6th 2009, 12:58 AM

#2  Physics Team
Join Date: Feb 2009
Posts: 1,425
 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, $\displaystyle u = \sqrt{2g(H38)}$…… (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) $\displaystyle \sqrt{2gH} = \sqrt{2g(H38)} + g.1$ . Solve for H with $\displaystyle g = 9.8 m/sec^2$
Last edited by physicsquest; Sep 6th 2009 at 10:13 AM.

 
Sep 6th 2009, 12:23 PM

#3  Junior Member
Join Date: Sep 2009
Posts: 2

wow.
i would've never thought of that.
thanks!

 
Sep 6th 2009, 02:22 PM

#4  Junior Member
Join Date: Sep 2009 Location: berkeley
Posts: 19

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

 
Sep 6th 2009, 02:28 PM

#5  Physics Team
Join Date: Jul 2009
Posts: 310

Originally Posted by nickerus 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.

 
Sep 6th 2009, 03:43 PM

#6  Physics Team
Join Date: Apr 2009 Location: Boston's North Shore
Posts: 1,548

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
$\displaystyle x_f = \frac{1}{2}gt_f^2$
$\displaystyle x_i = \frac{1}{2}gt_i^2$
$\displaystyle \Delta x = x_f  x_i = \frac{1}{2}g({t_f^2t_i^2})$
$\displaystyle t_f^2 = (t_i + 1)^2 = t_i^2 + 2t_i + 1$
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.

 
Sep 6th 2009, 03:49 PM

#7  Junior Member
Join Date: Sep 2009 Location: berkeley
Posts: 19

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:
$\displaystyle y(t1)=\frac{1}{2}(9.8)(t1)^2$
and
$\displaystyle y(t)=\frac{1}{2}(9.8)t^2$
we then say that
$\displaystyle y(t1)  38 = y(t)$
$\displaystyle (4.9)(t1)^2  38=(4.9)t^2$
$\displaystyle (4.9)(t^22t+1)=(4.9)t^2+38$
$\displaystyle t^22t+1=t^27.755$
$\displaystyle 2t=8.755$
$\displaystyle t \approx 4.3775$
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...
$\displaystyle \boxed{H \approx 93.90m}$

 
Sep 6th 2009, 03:57 PM

#8  Physics Team
Join Date: Jul 2009
Posts: 310

$\displaystyle \Delta x = V_it + \frac {1}{2}gt^2$
$\displaystyle V_i = \frac {\Delta x}{t}  \frac {1}{2}gt$
$\displaystyle V_i = 33.1 \mbox { m/s}$
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.
__________________
"Dissent is the highest form of patriotism."  Thomas Jefferson.
"Give me control of a nation's money and I care not who makes her laws." Mayer Amschel Rothschild
I study Mathematical Physics at the
University of Waterloo.
DC
Last edited by Deco; Sep 7th 2009 at 06:49 AM.

 
Sep 6th 2009, 04:40 PM

#9  Junior Member
Join Date: Sep 2009 Location: berkeley
Posts: 19

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.

 
Sep 6th 2009, 05:43 PM

#10  Physics Team
Join Date: Jul 2009
Posts: 310

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.)
__________________
"Dissent is the highest form of patriotism."  Thomas Jefferson.
"Give me control of a nation's money and I care not who makes her laws." Mayer Amschel Rothschild
I study Mathematical Physics at the
University of Waterloo.
DC
Last edited by Deco; Sep 6th 2009 at 06:16 PM.

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