SOLUTION: When Ms. Bibbi kicked a soccer ball, it traveled a horizontal distance of 150 feet and reached a height of 100feet at its highest point. Sketch the path of the soccer ball and find

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: When Ms. Bibbi kicked a soccer ball, it traveled a horizontal distance of 150 feet and reached a height of 100feet at its highest point. Sketch the path of the soccer ball and find      Log On


   

Question 875239: When Ms. Bibbi kicked a soccer ball, it traveled a horizontal distance of 150 feet and reached a height of 100feet at its highest point. Sketch the path of the soccer ball and find an equation of the parabola that models it.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
In physics class they teach you that the trajectory of an object is a parabola when
the object is close enough to Earth's surface,
it had some initial horizontal velocity,
and the only force acting on that object is gravity.

The path (with x as the horizontal distance and y as the height) looks like this:
graph%28300%2C300%2C-20%2C180%2C-0.1%2C0.9%2C-0.00015x%5E2%2B0.0225x%29 y%280%29=0 and y=150%29=0
And, of course, the maximum for y occurs halfway between those two points,
at x=%280%2B150%29%2F2=75 .
At that point, y%2875%29=100 .
As a second degree polynomial function with zeros at x=0 and x=150 ,
the function can be written, in factored form as
y=K%2A%28x-0%29%2A%28x-150%29 or y%28x%29=Kx%28x-150%29 .
To find the constant K we use the fact that y%2875%29=100 .
K%2A75%2875-150%29=100
K%2A75%28-75%29=100
-75%5E2%2AK=100
K=-100%2F75%5E2
That fraction can be simplified before we start multiplying away.

So the equation of the parabola is
y=-%284%2F225%29x%28x-150%29 ---> y=-%284%2F225%29x%5E2-150%28-4%2F225%29x ---> y=-%284%2F225%29x%5E2%2B%2840%2F15%29x .