Question 324529
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The height of an object in vertical motion near the earth's surface at time *[tex \Large t] is given by:


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


Where *[tex \Large v_o] is the initial velocity (assumed to be zero for this problem since the penny was dropped, not thrown) and *[tex \Large h_o] is the initial height, 70 feet in this case.


*[tex \LARGE \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +\ 70]


We want to know the time when *[tex \Large h(t)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ 16t^2\ =\ 70]


*[tex \LARGE \ \ \ \ \ \ \ \ t\ =\ \frac{\sqrt{70}}{4}]


We are ignoring the negative square root since it is unlikely that we will be able to make time go backwards.


The instantaneous velocity of a falling object at time *[tex \Large t] is given by:



*[tex \LARGE \ \ \ \ \ \ \ \ V(t)\ =\ -32t\ +\ v_o]


Clever calculus students will recognize this function as the first derivative of the height function.


All that is required is to calculate the velocity at the time calculated above, recalling that *[tex \Large v_o\ =\ 0]:


*[tex \LARGE \ \ \ \ \ \ \ \ V(t)\ =\ -32\frac{\sqrt{70}}{4}\ =\ -8\sqrt{70}]


You can push buttons on a calculator as well as I can.  "But wait," you say, "why is the velocity negative?"  Easy, the penny is going downward.


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