Question 521058
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First let's fix a couple of errors.  In the first place, *[tex \Large x] doesn't represent the vertical distance traveled.  *[tex \Large y] is the height and the height is on the vertical axis.  *[tex \Large x] represents the <i>horizontal</i> distance traveled.  Second, the point *[tex \Large \left(10,\,4\right)] is on the parabola but the point *[tex \Large \left(40,\,0\right)] most definately is NOT. *[tex \Large \left(40,\,4\right)] is on the parabola because *[tex \Large 0.5(40)\ -\ 0.01(40^2)\ =\ 20\ -\ 16\ =\ 4].


The rest is correct.


Now, for the last bit.  Parabolas are symmetrical about the axis which is a line through the vertex.  In this case, since the parabola is concave down, the axis is the vertical line *[tex \Large x\ =\ 25].  Since the parabola is symmetrical, the horizontal distance from the vertex back to the point where the projectile started, namely *[tex \Large \left(0,\,0\right)], is the same as the horizontal distance from the vertex to the point where the ball hits the ground again (and presumably, for this rather simplistic model anyway, the ball stops).  Since the ball started at a horiziontal distance of 0 and was at the vertex when it reached a horizontal distance of 25, it must have continued for another 25 feet before hitting the ground (0 height) and coming to rest.  Hence, the maximum horizontal travel was 25 + 25 = 50 feet.


You might care to notice that if you were to set your function equal to zero and solve the quadratic, you would have two roots, namely 0 and 50.


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