Question 196347
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t) = -16t^2 + h_o]


but you are given that *[tex \LARGE h_o = 244], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t) = -16t^2 + 244]


You want to know the value of *[tex \LARGE t] when *[tex \LARGE h(t) = 0], presuming you are considering the ground to be at a height of zero.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -16t^2 + 244 = 0]


from which you obtained:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2 = \frac{244}{16} = \frac{61}{4}]


Simply take the square root of both sides.  Under ordinary circumstances, you would consider both the positive and negative square roots.  However, in this case the negative answer is absurd because the object most certainly did not hit the ground at some time <i>before</i> it was dropped.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \sqrt{\frac{61}{4}} = \frac{\sqrt{61}}{\sqrt{4}} = \frac{\sqrt{61}}{2}]


61 being prime, this is the exact answer reduced to simplest terms.  You can punch it in to a calculator to get a numerical approximation if you like -- should be right around 3.9 or so.


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