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 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, $\displaystyle u = \sqrt{2g(H-38)}$…… (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.  Tags falling, object Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post jlyu002 Kinematics and Dynamics 5 Jul 15th 2014 10:28 AM mryeh Electricity and Magnetism 5 Apr 4th 2014 05:54 AM Deadstar Kinematics and Dynamics 2 Apr 5th 2010 12:44 PM tweedydaf Kinematics and Dynamics 2 Sep 18th 2008 10:33 PM colombo Kinematics and Dynamics 1 Sep 6th 2008 09:41 PM