Question 282442
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Unless there is some sort of anti-gravity machine attached to this wrench, then your expression for height as a function of time is incorrect.  The physics of objects near the earth surface as I understand them demands that the function be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S(t)\ =\ -16t^2\ -\ 32t\ +\ 128]


Now, to answer part a, read the problem again, specifically this part:  "the height of the wrench above the ground after *[tex \Large t] seconds is given by:"  Let's read that again: "the height of the wrench above the ground after *[tex \Large t] seconds is given by:"  So doesn't it make sense that the height of the wrench above the ground after 1 second would be given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S(1)\ =\ -16(1)^2\ -\ 32(1)\ +\ 128] ?


Just do the arithmetic.


Part b).  The ground is height 0.  So if:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S(t)\ =\ 0]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ -\ 32t\ +\ 128\ =\ 0]


Just solve the quadratic -- it factors quite neatly, in fact.  Hint:  Divide through by -16 first.  As with any quadratic, you will get two roots.  You can discard the negative one because the wrench certainly didn't hit the ground some number of seconds before the guy threw it.  Your answer is the positive root.


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