. An object thrown from a height of 2 m above the ground follows a parabolic path until the object falls to the ground. If the object reaches a maximum height (measured from the ground) of 7 m after travelling, a horizontai distance of 4 m, determine the horizontal distance between the object's initial and final positions.

We can solve this problem by modeling the motion of the object as a parabola in the \(xy\)-plane. Given the information, we will assume that the object follows a path described by the equation of a parabola.

### Step 1: General form of a parabola
The equation for the parabolic path of the object can be written as:
\[
y = ax^2 + bx + c
\]
where:
- \(y\) is the height (vertical position) of the object at a given horizontal position \(x\),
- \(a\), \(b\), and \(c\) are constants.

### Step 2: Initial conditions and vertex of the parabola
We are given the following points:
- At \(x = 0\), the height of the object is 2 meters (initial position). Therefore, \(y(0) = 2\), which gives \(c = 2\).
\[
y(0) = 2 \implies c = 2
\]
- At \(x = 4\) meters, the height is 7 meters (maximum height). This means that the vertex of the parabola occurs at \(x = 4\), and the corresponding height is 7 meters.

So, the equation now becomes:
\[
y = ax^2 + bx + 2
\]

### Step 3: Using the vertex form
The vertex form of a parabola is:
\[
y = a(x - h)^2 + k
\]
where \((h, k)\) is the vertex. For our case, the vertex is at \((4, 7)\), so the equation becomes:
\[
y = a(x - 4)^2 + 7
\]

### Step 4: Applying boundary condition at \(x = 0\)
We know that at \(x = 0\), \(y = 2\). Substituting this into the equation to find \(a\):
\[
2 = a(0 - 4)^2 + 7
\]
\[
2 = a(16) + 7
\]
\[
a = \frac{2 - 7}{16} = -\frac{5}{16}
\]

### Step 5: Parabola equation
Thus, the equation of the parabolic path is:
\[
y = -\frac{5}{16}(x - 4)^2 + 7
\]

### Step 6: Determine where the object hits the ground
To find where the object hits the ground, set \(y = 0\) (since the object lands on the ground) and solve for \(x\):
\[
0 = -\frac{5}{16}(x - 4)^2 + 7
\]
\[
\frac{5}{16}(x - 4)^2 = 7
\]
\[
(x - 4)^2 = \frac{112}{5} = 22.4
\]
\[
x - 4 = \pm \sqrt{22.4}
\]
\[
x - 4 = \pm 4.733
\]
Thus, the two solutions for \(x\) are:
\[
x = 4 + 4.733 = 8.733 \quad \text{or} \quad x = 4 - 4.733 = -0.733
\]

### Step 7: Horizontal distance between initial and final positions
The object was thrown from \(x = 0\) and lands at \(x = 8.733\). Therefore, the total horizontal distance between the object's initial and final positions is:
\[
\text{Horizontal distance} = 8.733 - 0 = 8.733 \, \text{m}
\]

Thus, the horizontal distance between the object's initial and final positions is approximately **8.73 meters**.

Would you like more details on any specific step? Here are some related questions:

1. How do you derive the equation for a parabolic path in projectile motion?
2. Can you explain how to find the vertex form of a parabola?
3. What is the significance of the vertex in a projectile motion problem?
4. How can we use boundary conditions to determine constants in an equation?
5. What role does the initial height play in solving projectile motion problems?

**Tip:** In projectile motion, the horizontal and vertical motions are independent. The horizontal distance is determined by the time of flight, while the vertical motion follows a parabolic trajectory.

An object thrown from a height of 2 m above the ground follows a parabolic path until the object falls to the ground. If the object reaches a maximum height (measured from the ground) of 7 m after travelling a horizontal distance of 4 m, determine the horizontal distance between the object's initial and final positions.

Discover how to determine the horizontal distance covered by an object thrown from a height of 2 meters following a parabolic path. This problem explores key concepts in algebra, quadratic equations, and projectile motion, providing a step-by-step solution for students in Grades 9-11.

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