Question 33150This question is from textbook College Algebra
: A ball is thrown across a playing field. Its path is given by the equation y= -0.005x^2 +x +5, where x is the distance the ball has traveled horizontally, and y is its height above ground level, both measured in feet.
a. What is the maximum height attained by the ball?
b. How far has it traveled horizontally when it hits the ground?
I know this involves vertex and quadratic equations but it doesn't graph right.
Thank you very much!
This question is from textbook College Algebra
Answer by mukhopadhyay(490)   (Show Source):
You can put this solution on YOUR website! y=-0.005x^2+x+5 is a quadratic function representing parabola opening downward.
The maximum height is reached at its vertex; Hence the y-coordinate of the vertex will give us the required maximum height the ball can attain.
If there is an x-intercept of a vertex, the larger x-intercept will give us the the distance the ball would travel horizontally when it hits the ground;
- I am providing the above explanation so that you will not only be able to solve this type of problems, but also be able to develop a concept behind this-......
x-coordinate of the vertex is attained at -b/2a (for function ax^2+bx+c);
Here, a=-.005; b=1; Hence x-coordinate of vertex = -1/(2)(-.005) = 1/.01 = 100;
y-coordinate of the vertex is f(-b/2a)
= f(100) = -.005(10^4) + 100 + 5 = -50+105 = 55;
Answer (1): Maximum height the ball can attain is 55 feet;
.........
x-intercepts are found by equating f(x) = 0;
So, -0.005x^2+x+5 = 0
=> .005x^2-x-5 = 0
=> 5x^2-1000x-5000 = 0
=> x^2-200x-1000 = 0
=> larger x= {-(-200) + Sqrt[(-200)^2-4(1)(-1000)]]/2(1) = [200+sqrt(44000)]/2
= 100+55sqrt(10);
Answer (2): The distance the ball would travel horizontally when it hits the ground is 100+55*sqrt(10) feet.
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