Question 247789
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ 16t^2]


But you are given *[tex \Large s] and need to determine *[tex \Large t]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16t^2\ =\ s]


Multiply both sides by *[tex \Large \frac{1}{16}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2\ =\ \frac{s}{16}]


Take the square root of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \pm\sqrt{\frac{s}{16}}\ =\ \pm\frac{\sqrt{s}}{4}].


Discard the negative root; the idea of negative time in this context being an absurdity.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\sqrt{s}}{4}]


To obtain your specific answer, substitute the given value for *[tex \Large s] and then do the required arithmetic.


By the way, that is an unbelievably tall tree.  I say unbelievably because "Lost Monarch," the tallest of the California Coastal Redwoods and said to be the world's tallest tree is approximately 320 feet tall measured from the ground.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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