Question 275603
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Your height function is improperly stated.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S(t)\ =\ \frac{1}{2}gt^2\ +\ v_ot\ +\ s_o]


Where *[tex \Large g] is the acceleration in feet per second due to gravity, namely *[tex \Large -32\text{ \frac{ft}{sec^2}]


and *[tex \Large v_o] is the initial velocity, here *[tex \Large -32\text{ \frac{ft}{sec}]


and *[tex \Large s_o] is the initial height, here 128 feet.


Hence:


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


The height of the wrench after 1 second is the value of the function when *[tex \Large t\ =\ 1].  Plug in the value and do the arithmetic:


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


The height of the ground is zero, so the time value when *[tex \Large S(t)\ =\ 0] is the length of time it takes for the wrench to hit the ground -- presuming that the wrench didn't hit something else, perhaps the worker's boss' head -- before it hit the ground.  So, solve the quadratic:


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


Hint: First factor out -16


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