Physics Help Forum Falling object

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 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, $\displaystyle u = \sqrt{2g(H-38)}$…… (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.
 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
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 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: 1,576 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.
 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: $\displaystyle y(t-1)=\frac{1}{2}(-9.8)(t-1)^2$ 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... $\displaystyle \boxed{H \approx 93.90m}$
 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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