Question 312082
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We are given that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ -16t^2\ -\ 2t\ +\ 524]


So we are at the top of a 524 foot tall building throwing a rock downward with an initial velocity of 2 feet per second toward the ground.


But we want to know the time when


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 321] feet.


Set the two expressions equal to each other because they are both equal to  *[tex \LARGE d].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ -\ 2t\ +\ 524\ =\ 321]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ -\ 2t\ +\ 203\ =\ 0]


Just solve the quadratic for *[tex \LARGE t].  Hint:  The quadratic factors because 2 times -8 is -16, (-8 times -7) plus (2 times -29)\ = 56 minus 58 = minus 2, and -7 times -29 is 203.


One of the roots is negative -- discard it.  You don't care what happened <b><i>before</i></b> you threw the rock.  The positive root is the number of seconds that will have elapsed when the height of the rock is 321 feet. 


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