SOLUTION: A cannon fires a cannonball as shown in the figure. The path of the cannonball is a parabola with vertex at the highest point of the path. If the cannonball lands d = 600 ft from t

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A cannon fires a cannonball as shown in the figure. The path of the cannonball is a parabola with vertex at the highest point of the path. If the cannonball lands d = 600 ft from t      Log On


   

Question 374241: A cannon fires a cannonball as shown in the figure. The path of the cannonball is a parabola with vertex at the highest point of the path. If the cannonball lands d = 600 ft from the cannon and the highest point it reaches is h = 1200 ft above the ground, find an equation for the path of the cannonball. Place the origin at the location of the cannon.
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Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A cannon fires a cannonball as shown in the figure.
The path of the cannonball is a parabola with vertex at the highest point of the path.
If the cannonball lands d = 600 ft from the cannon and the highest point it
reaches is h = 1200 ft above the ground, find an equation for the path of the cannonball.
Place the origin at the location of the cannon.
:
Since the cannon ball starts at x=0 and lands x=600, the axis of symmetry = 300
:
Using the form ax^2 + bx + c = y, (c=0 in this problem)
x=300; y=1200
(300^2)a + 300b = 1200
90000a + 300b = 1200
Simplify, divide by 300, results
300a + b = 4
:
x=600; y= 0
36000a + 600b = 0
Simplify, divide by 600
600a + b = 0
:
Use elimination here
600a + b = 0
300a + b = 4
---------------subtraction eliminates b, find a
300a = -4
a = %28-4%29%2F300 = -1%2F75
:
Find b using 300a + b = 4
300(-1/75) = b
-4 + b = 4
b = +8
:
The equation: h(x) = -1%2F75x^2 + 8x
:
Graphically:
+graph%28+300%2C+200%2C+-200%2C+800%2C+-300%2C+1500%2C+%28-1%2F75%29x%5E2%2B8x%29+
You can see the vertex, x=300, y=1200