Question 330635
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The first thing I need to do is to correct your height function.  The gravitational acceleration term is based on the square of time, not the square of the gravitation coefficient. Furthermore, I like to work with my polynomials in standard form, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t,v_o,h_o)\ =\ -4.9t^2\ +\ v_ot\ +\ h_o]


All you need to do is plug in the given numbers and do the indicated arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(3,30,110)\ =\ -4.9(3)^2\ +\ (30)(3)\ +\ (110)]


I'll leave you alone so you can spend some quality time with your calculator.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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