Question 389681
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*[tex \LARGE\ \ \ \ \ \ \ \ \ \ h(t)\ =\ c\ -\ \left(d\ -\ 4t\right)^2]


Expand the binomial:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +\ 8dt\ +\ c\ -\ d^2]


Compare this to the standard form of the height function for a projectile near the surface of the Earth with respect to time:


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


By comparison you can see that the initial height, *[tex \Large h_o] , must be equal to *[tex \Large c\ -\ d^2], and the initial velocity, *[tex \Large v_o\ =\ 8d].


We know that the maximum height is obtained at the time equal to the *[tex \Large t]-coordinate of the vertex of the parabola that is the graph of the function.  The *[tex \Large t]-coordinate is found by:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ t_{max}\ =\ \frac{-8d}{-32}]


but we were given that *[tex \Large t_{max}\ =\ 2.5]


From which we can deduce:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ d\ =\ 10]


(verification left as an exercise for the student)


and then we can determine that


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ v_o\ =\ 80]


Next we were given that *[tex \Large h_o\ =\ 6], but knowing that *[tex \Large d\ =\ 10\ \rightarrow\ d^2\ = 100], we can deduce that:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ c\ =\ 106]


Plug in the values:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ h(t)\ =\ 106\ -\ \left(10\ -\ 4(1)\right)^2]


And then do the arithmetic.


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