Question 1177200
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Your units are feet and seconds, so distance in feet, time in seconds, the magnitude of velocity is feet per second, and magnitude of the acceleration is feet per second per second.


The presumption here is that the building in the story is on the surface of planet Earth where the acceleration due to gravity is *[tex \Large g\ =\ -32\text{ft/sec^2}] and motion downward, i.e. toward the center of the earth, is indicated by negative acceleration and velocity magnitudes.


The height as a function of time for an object traveling vertically with respect to the earth's surface is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  h(t)\ =\ \frac{1}{2}gt^2\ +\ v_ot\ +\ h_o]


Where *[tex \Large g] is the acceleration due to the force of gravity, *[tex \Large v_o] is the initial velocity magnitude (recall that the velocity magnitude must be a negative quantity for an object thrown downward), and *[tex \Large h_o] is the initial height.


The instantaneous velocity as a function of time for this situation is the first derivative of the height function, namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v_i\ =\ \frac{dh}{dt}\|_{t\,=\,t_i}\ =\ -32t_i\ +\ v_o]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v_2\ =\ \frac{dh}{dt}\|_{t\,=\ 2}\ =\ -32(2)\ -\ 30\ \text{ft/sec}]


You can do your own arithmetic.

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