Question 1162345
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The general function for the height of a projectile launched vertically as a function of time is:


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


Where *[tex \Large g] is the acceleration resulting from the force of gravity, either *[tex \Large -32\text{ ft/sec^2}] or *[tex \Large -9.8\text{ m/sec^2}] depending on whether you are using fps or mks units, *[tex \Large v_o] is the initial velocity, *[tex \Large h_o] is the height at launch, and *[tex \Large t] is the elapsed time in seconds since launch.


Given your lead coefficient of -16, I'll assume fps.


Assuming the referee allowed the count to hit the ground rather than catching it as is typical behavior at the start of football games, then the coin will be in the air from the time of launch until it hits the ground.  When it hits the ground its height above the ground is necessarily zero.


So solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ +\ 70t\ +\ 9\ =\ 0]


Note that since the initial height is more than zero, the two roots of this equation must be of opposite signs.  You want the positive answer because what happened before he threw the coin is moot.


As a side note, if the referee was able to reach 9 feet in the air standing flat-footed, he would have to be about 6 feet 9 inches tall on average.  6 feet would be a more reasonable value for initial height.  Who reaches above their head to release the coin when tossing it?
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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