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anonymous
 one year ago
At the National Physical Laboratory in England, a measurement of the freefall acceleration g was made by throwing a glass ball straight up in an evacuated tube (the air was pumped out) and letting the ball return. Let ΔTL in the figure below be the time interval between the two passages of the ball across a certain lower level, ΔTU the time interval between the two passages across an upper level, and H the distance between the two levels.
What is g in terms of the measurements taken?
anonymous
 one year ago
At the National Physical Laboratory in England, a measurement of the freefall acceleration g was made by throwing a glass ball straight up in an evacuated tube (the air was pumped out) and letting the ball return. Let ΔTL in the figure below be the time interval between the two passages of the ball across a certain lower level, ΔTU the time interval between the two passages across an upper level, and H the distance between the two levels. What is g in terms of the measurements taken?

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442819505362:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442825943411:dw setting the coordinates as shown to simplify algebra and using equation of motion \(x = ut + \frac{1}{2}at^2 = t(u+\frac{1}{2}at)\), we can say \(\large 0 = t(u+\frac{1}{2}at) \implies t_{o1} = 0, t_{o2} =\Delta T_L = \frac{2u}{a}\) we can also say \(\large H = ut + \frac{1}{2}at^2 \implies at^2 +2ut  2H = 0\) and from the quadratic formula we now know that \(\large \Delta T_U = 2 \times \frac{\sqrt{(2u)^2  (4)(a)(2H)}}{2a} \) but that's an ugly way to do it, especially given how neat it suggests the solution is.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442827632092:dw exploiting this symmetry might provide a more fun solution
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