Question 314958
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No can do.  If you use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ 16t^2\ +\ vt\ +\ s]


presuming *[tex \LARGE v] is the initial velocity and *[tex \LARGE s] is the initial height, then this frog must have a big red S emblazoned on its chest because once this frog takes off he is NEVER, and I repeat NEVER, going to come down again.


The reason I am so certain of this is that you have the factor that accounts for the acceleration due to gravity (the *[tex \LARGE 16t^2] part) the same sign as the initial velocity part.  So if the frog is jumping into the air with a positive initial velocity and the acceleration due to gravity is also positive, then what makes the frog come back down again?


Write back after you verify that the appropriate equation for height as a function of time is:


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


where *[tex \Large v_o] is the initial velocity and *[tex \Large h_o] is the initial height.


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