Question 183961
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*[tex \LARGE \text{          }\math h(t)=-16t^2+48t+74]


(By the way, this is true if and only if the height of the building plus the height of Becky's hand above the top of the building at the time she released the ball was 74 feet and she threw the ball with an initial velocity of 48 feet per second.)


Your task is to determine the value of <i>t</i> that makes <i>h(t)</i> equal to 10.


So:


*[tex \LARGE \text{          }\math h(t)=-16t^2+48t+74 = 10]


Put the quadratic in standard form:


*[tex \LARGE \text{          }\math -16t^2+48t+64 = 0]


Multiply by *[tex \Large -\frac{1}{16}] :


*[tex \LARGE \text{          }\math t^2 - 3t - 4 = 0]


Now all you need to do is factor and solve for <i>t</i>.  You will get two roots to the quadratic, as we would expect, but one of them will be negative.  Since the negative root leads to the absurdity that the ball was at 10 feet off the ground at some time <i><b>before</b></i> she threw the ball, exclude that root as extraneous.  The positive root will be the desired time value.


Hint:


*[tex \LARGE \text{          }\math (-4) + (1) = -3]


and


*[tex \LARGE \text{          }\math (-4)(1) = -4]


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