Question 1165049
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Evaluate the height function, *[tex \Large h(t)\ =\ -4.9t^2\ +\ v_ot\ +\ h_o] where *[tex \Large v_o\ =\ -5.5\text{ m/s}] and *[tex \Large h_o] is the height of the building.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(1.70)\ =\ -4.9(1.70)^2\ -\ 5.55(1.70)\ +\ 50]


This gives you the height at the specified time.  As long as this value is less than the height of the building, subtract the height at that time from the height of the building to determine the distance fallen.


Instantaneous velocity is the derivative of the height function evaluated at the time when the height is zero.


Solve the quadratic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4.9t^2\ -\ 5.55t\ +\ 50\ =\ 0]


to find the elapsed time when the balloon hits the ground, *[tex \Large t_f].


 
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -4.9t^2\ -\ 5.55t\ +\ 50]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dt}\,h(t)\ =\ -9.8t\ -\ 5.55]


Evaluate the derivative at *[tex \Large t_f]

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