Question 461455: I hope this is the right category. I wasn't sure where to put it. The beginning intro to the problem is kinda long. Sorry about that.
Ignoring air-resistance, the path of a projectile (a cannonball, a tight wad of paper), when close to the earth's surface, is approximately parabolic and is therefore described by a quadratic function. If the cannonball is launched at the point (0,0) in the xy-plane, its height h in feet as a function of x, the horizontal distance in feet from the launch point is given by
h = f(x) =
-16(1+m^2)x^2+mx
.....------
......v^2o
(The little o, I'm not sure how to type that in. My paper shows v squared and then a teeny zero to the right and lower. Sorry. It's been about 19 years since I've taken Algebra, so I'm not up on the language. It was also hard to type the fraction, but the actual fraction is (1+m^2) over v^2o. The rest isn't written on the fraction line but is beside the others to multiply the fraction by. Please ignore the periods.)
In this formula, the symbol vo (again, that little 0) is the initial launch speed in ft/sec and m is the slope of the initial trajectory. For example, when m=1, the cannon in our sketch makes an angle of 45 degrees with the horizontal. (Recal a line of slope 1 makes an angle of 45 degrees with the x-axis).
If m=1, and vo=80 ft/sec, you can check for yourself that this formula simplifies to:
y = h(x) =
-16(1+1^2)x^2+1x = -0.005x^2+x
......------
......80^2
Now, here is the problem (finally):
Use your knowledge of quadratic equations to find the following quantities:
* How far away from the launch point the ball hits the ground; this is called the range.
* the maximum height of the ball while it's in the air.
If anyone can help me understand how to even go about answering this, I would be beyond appreciative. I have read and read and re-read this problem, and I feel like I still have no clue what to do. And the scary thing is that this is only the first part of the question.
Thank you!
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! the "little o" is for "v zero" (the velocity at time zero, or initial velocity)
using the parabola that describes the trajectory, solve for the zeros (roots)
___ one root will be the starting (launch) point, and the other root will be the end point (range)
___ h(x) equals zero at the beginning and the end of the projectile flight
0 = -0.005x^2 + x ___ factoring ___ 0 = x(-0.005x + 1)
0 = x ___ this is the launch
0 = -0.005x + 1 ___ 0.005x = 1 ___ x = 200 ___ this is the range
the maximum height of the projectile occurs at the vertex of the parabola, on the axis of symmetry
___ the general equation is ___ x = -b / (2a)
in this case ___ x = -1 / [2(-0.005)] ___ x = 100 (this is half the range, as would be expected)
___ use this x-value to find the max height
h(100) = -0.005(100)^2 + (100) ___ h(max) = -50 + 100 ___ h(max) = 50
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