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
1600 ft from the cannon and the highest point it reaches is
3200 ft above the ground, find an equation for the path of the
cannonball. Place the origin at the location of the cannon.
A) y-800=(-1/160)(x-4000)^2
B) y-800=(-1/120)(x-4000)^2
C) y-4000=(-1/160)(x-800)^2
D) y-4000=(-1/120)(x-800)^2
From your description, the vertex must be (800, 3200)
so its equation must be
y = a(x-800)^2 + 3200
but (1600,0) also lies on it, so
0 = a(1600-800)^2 + 3200
a = -3200/(800)^2 = - 1/200
the equation would be y = (-1/200)(x - 800)^2 + 3200
or in the form they use:
y - 3200 = (-1/200)(x - 800)^2
none of the choices match that.
To find the equation of the parabolic path, we need to determine the equation in the form y = ax^2 + bx + c, where (x, y) represents a point on the path.
1. The vertex of the parabola represents the highest point in the path. In this case, the vertex is at (0, 3200).
2. We also know that the cannonball lands 1600 ft from the cannon, which means it will reach this point at some x-coordinate value. Let's call this value x1.
3. Since the vertex is at (0, 3200) and the cannonball lands 1600 ft away, the x-coordinate of the landing point is 1600.
4. To find the equation, we'll use the vertex form of a parabola: y = a(x-h)^2 + k, where (h, k) represents the vertex.
Substituting the known values, we have:
y = a(x-0)^2 + 3200
Simplifying further:
y = ax^2 + 3200
Now, we need to find the value of a. Using the landing point (1600, 0), we can substitute these values into the equation:
0 = a(1600)^2 + 3200
Simplifying:
0 = 2560000a + 3200
Solving for a:
2560000a = -3200
a = -3200 / 2560000
a = -1 / 800
Therefore, the equation for the path of the cannonball is:
y = (-1/800)x^2 + 3200
Comparing this with the answer choices, we can see that the correct equation is B) y - 800 = (-1/120)(x - 4000)^2.
To find an equation for the path of the cannonball, we can make use of the equation for a parabola in vertex form:
y - k = a(x - h)^2
where (h, k) represents the coordinates of the vertex of the parabola and 'a' determines the shape and direction.
In this case, the vertex of the parabola represents the highest point the cannonball reaches, which is given as (0, 3200) since we've placed the origin at the cannon. This allows us to substitute h = 0 and k = 3200 into the equation, giving us:
y - 3200 = a(x - 0)^2
y = a(x^2)
Now, we need to find the value of 'a'. We have the information that the cannonball lands 1600 ft from the cannon, which means it lands at the point (1600, 0). Substituting these values into the equation gives us:
0 = a(1600^2)
0 = 2560000a
Since a cannot be zero (as it determines the shape of the parabola), we can solve for 'a':
2560000a = 0
a = 0
Therefore, the equation for the path of the cannonball is y = 0, which represents a straight line parallel to the x-axis. None of the given answer choices match this equation, so the correct option is not listed.