SOLUTION: The trajectory of the ball is modelled on the equation y= -x^2/2+3x+8 I need to find how far the ball will be horizontally from the house when it first lands on the ground after b

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The trajectory of the ball is modelled on the equation y= -x^2/2+3x+8 I need to find how far the ball will be horizontally from the house when it first lands on the ground after b      Log On


   

Question 473676: The trajectory of the ball is modelled on the equation y= -x^2/2+3x+8
I need to find how far the ball will be horizontally from the house when it first lands on the ground after being thrown from the balcony.
I,m sure this is very simple to answer but I seem unable to do
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The trajectory of the ball is modeled on the equation y = -x%5E2%2F2 + 3x + 8
this means y = the height of the ball at distance x from the house
:
When the ball strikes the ground, h = 0, find x, therefore
-x%5E2%2F2 + 3x + 8 = 0
we can make this easy to factor by multiplying equation by -2, results
x^2 - 6x - 16 = 0
Factors easily to
(x-8)(x+2) = 0
the positive solution is all we want here
x = 8 units is the distance from the house
;
;
A graph of the original equation will illustrate this:
+graph%28+300%2C+200%2C+-4%2C+12%2C+-4%2C+16%2C+%28-x%5E2%2F2%29%2B3x%2B8%29+
note that
when x=0 the ball is at height of 8 which is where it is thrown from the house
when y=0 the ball strikes the ground 8 units from the house