Question 1163928
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Using the mks measurement system, the function that describes the height of a projectile where the gravitational acceleration constant is *[tex \Large g] at time *[tex \Large t] with an initial vertical velocity component magnitude of *[tex \Large v_o] and an initial height of *[tex \Large h_o] is:

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


For objects near the earth's surface, *[tex \Large g\ =\ -9.8\text{ m/sec^2}]


Since you don't specify the height of the ball when the bat impacted it, we have to assume *[tex \Large h_o\ =\ 0]


That makes your function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -4.9t^2\ +\ 42t]


This would graph as a concave down parabola because the lead coefficient is negative.  The answer to part a) is the value of *[tex \Large t] at the vertex of the parabola.  The answer to part b) is the value of the function at the answer to part a.


If you have a parabola described by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(t)\ =\ at^2\ +\ bt\ +\ c]


The *[tex \Large t]-coordinate of the vertex is given by *[tex \Large -\frac{b}{2a}]


The value of the function at the vertex is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho\(-\frac{b}{2a}\)\ =\ a\(-\frac{b}{2a}\)^2\ +\ b\(-\frac{b}{2a}\)\ +\ c]


4.3 seconds is close enough for part a, but part b is inaccurate by an amount equal to the actual height of the ball when the bat hit it.  The bat certainly wasn't on the ground if the batter hit a pop-up.  So, that bat had to be a meter to a meter and a half off of the ground -- perhaps more, and the real world answer to part b would be somewhere in the range of 91 to 92 meters.

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