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The height of a baseball is modeled by the function $h(x)=-0.005x^2+0.3x+1.5$, would an outfielder which is modeled by the function $m(x)=-0.06x+5.6$ where $50 \le x \le 90$, catch the ball?

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ive made m(x)=h(x) and found the discriminant of that but i dont know where to go from there – jasminder88 Apr 1 '13 at 2:23
Are you sure you wrote the problem correctly? Also, can you tell us your thoughts on the problem so we can provide better guidance and help resolve the issue? Regards – Amzoti Apr 1 '13 at 2:23
Does $x$ in this case represent time or distance? – apnorton Apr 1 '13 at 2:26
when m(x)=h(x) the equation is $-0.005x^2+0.36x-4.1$ , after this i found the discriminant and it was greater than 0 so that means that there are two solutions for x – jasminder88 Apr 1 '13 at 2:26
distance from the homebase, sorry about that – jasminder88 Apr 1 '13 at 2:26
up vote 3 down vote accepted

Answer to your question:

Look at this graph of the equations. The blue represents the height of the baseball, and the red represents something about the outfielder. Notice one of the solutions is well outside the accepted range for $x$.

Some things that don't sit right with me:

Firstly, the height of a baseball is governed (without considering air resistance) by the equation $y = -\frac{g}{2}t^2 + v_0t + y_0$, where $g$ is the gravitational acceleration. So, this occurs on some planet (or with some units) where gravitational acceleration is $-0.005$. Whoever set the problem should have attached some units, or used an accurate gravitational acceleration.

Also, the process of setting the expressions equal to each other is very wrong. One function returns the elevation of the baseball, the other is the position of the outfielder. In order for setting them equal to make sense, we either need the function for the outfielder to return the outfielder's elevation with respect to time (which is very odd, to say the least). Or, we can have the function for the baseball return its distance in the horizontal plane. This makes more sense, but is against the problem definition.

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