Question 1027440
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The path of a thrown or batted ball is approximately a parabola.  It would have to be an indoor baseball stadium with the air evacuated to a perfect vacuum to make it a perfect parabola, but a parabola is close enough for this problem.


Since we are given the coordinates of the vertex of the parabola, a good place to start is the vertex form of an equation of a parabola, namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ =\ a(x\ -\ h)^2\ +\ k]


where *[tex \Large (h, k)] is the vertex.  Since your vertex is *[tex \Large (210,60)], we can directly determine:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ =\ a(x\ -\ 210)^2\ +\ 60]


Since the parabola also contains the point *[tex \Large (0,3)], it must be true that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(0\ -\ 210)^2\ +\ 60\ =\ 3]


From which you can calculate the value of *[tex \Large a] to complete the derivation of the desired equation.  In that this step is purely arithmetic, I will leave it up to you.  Hint:  Since perforce this parabola must open downward, i.e. the function value at the vertex is a maximum, it must be true that *[tex \Large a\ <\ 0].  If you do not get a negative result for your calculation, you have made an error.


Once you know the value of *[tex \Large a], you can find the function value at *[tex \Large x\ =\ 400].  The amount that this function value is greater than 10 is the amount by which the ball clears the outfield fence, or the amount by which the function value is less than 10 is the amount by which the ball fails to clear the fence.


Assuming that there is nothing more than level ground on the outside of the outfield fence and the ball does, indeed clear the fence, then the domain is the set of real numbers in the interval *[tex \Large (0, x_{max})].  More on the value of *[tex \Large x_{max}] later.


On the other hand, if the ball does NOT clear the fence, the domain is the range *[tex \Large (0,400)].  We don't care what happened before the ball was hit and we don't care what happens after the ball hits the fence.


If the ball does not clear the fence, then the range is the interval *[tex \Large (y_{min},60)].  *[tex \Large y_{min}] is either 3 feet or the height above the ground that the ball hits the fence or zero if the ball hits the ground before it reaches the fence (if your calculation to part b) reveals that the amount the ball fails to clear the fence is greater than 10 feet.) whichever is the smallest.


If the ball clears the fence and we are assuming level ground outside the fence, then the range is the interval (0,60).


Now, as to the value of *[tex \Large x_{max}].  If the ball clears the fence, this value is 400 feet plus the distance that the ball travels before it hits the ground on the other side of the fence.


Using the value of *[tex \Large a] calculated in part a) of the problem, solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a(x\ -\ 210)^2\ +\ 60\ =\ 0].


Hint:  Expand the binomial, collect like terms, and then use the quadratic formula.  One root will be some small negative number (where the parabola would meet the ground behind the batter), and the other will be *[tex \Large x_{max}].


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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