Question 186049
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Your height function is incorrect.  What you have is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t) = 60t - (16t)(16t) = 60t - 256t^2]


The correct function, in the form you presented is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t) = 60t - 16(t)(t)]


Or, in a more traditional form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t) = -16t^2 + 60t]


Now, what you need to know is the value of <i>t</i>, different from 0 (because time 0 is when it left the ground in the first place), when the height will be zero.  So just set the right side equal to zero and solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2 + 60t = 0]


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t(-16t + 60) = 0]


Use the Zero Product Rule to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = 0]


Which we exclude as previously discussed, or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t + 60 = 0]


I'll let you finish.  Remember to reduce any fractions to lowest terms.


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