A firework is projected from the lowest corner of a pole of height 
15
15m into the air and reaches a maximum height which is same as the height of the pole. The path of the firework is parabolic and it lands 
30
30m from the launch site 
B
B (as shown in the figure). A ball is thrown from the top of the pole which follows the parabolic path (whose vertex is at the point of the throw (
A
A)) and lands at a distance of 
20
20m from the ground (as shown in the figure). Taking 
B
B as the origin, horizontal direction as the 
X
X axis, vertical direction as the 
Y
Y axis, and unit length to be 
1
1m, answer the question 1, 2 and 3.



To analyze the problem, let's break it down step by step using the information provided:

### Given Information:

1. **Firework Path:**
   - The firework is launched from point \( B \), which is the origin \( (0, 0) \).
   - It reaches a maximum height of \( 15 \) meters, which is the same height as the pole.
   - It lands \( 30 \) meters from the launch site (point \( B \)).
   - The path of the firework is parabolic.

2. **Ball Path:**
   - The ball is thrown from the top of the pole (point \( A \)), which is at \( (0, 15) \).
   - The ball lands at a distance of \( 20 \) meters horizontally from the point of throw.
   - The path of the ball is also parabolic.

### Question 1: **Equation of the Firework's Path**

The path of the firework is a parabola. Since it starts at the origin \( (0, 0) \) and the vertex is at the point \( (15, 15) \), the equation of the parabola can be expressed in vertex form:

\[
y = ax^2 + bx + c
\]

Since the parabola passes through the origin \( (0, 0) \), \( c = 0 \), so the equation simplifies to:

\[
y = ax^2 + bx
\]

The parabola also reaches a maximum height of \( 15 \) meters at \( x = 15 \). Using the vertex form, we know that the general form for a parabola with a vertex at \( (h, k) \) is:

\[
y = a(x - h)^2 + k
\]

Here, \( h = 15 \) and \( k = 15 \). Thus, the equation becomes:

\[
y = a(x - 15)^2 + 15
\]

We also know that the firework lands \( 30 \) meters from the launch site at \( x = 30 \). At this point, \( y = 0 \), so:

\[
0 = a(30 - 15)^2 + 15
\]

Solving this for \( a \):

\[
-15 = a(15)^2
\]

\[
a = -\frac{15}{225} = -\frac{1}{15}
\]

Therefore, the equation of the firework's path is:

\[
y = -\frac{1}{15}(x - 15)^2 + 15
\]

### Question 2: **Equation of the Ball's Path**

The ball is thrown from the top of the pole at \( (0, 15) \). The parabola's vertex is at \( (0, 15) \), and it lands \( 20 \) meters away. The equation of the parabola can be written as:

\[
y = ax^2 + bx + c
\]

Given that the vertex is at \( (0, 15) \), the equation simplifies to:

\[
y = ax^2 + 15
\]

Since the ball lands \( 20 \) meters away (at \( x = 20 \)), and \( y = 0 \) at this point:

\[
0 = a(20)^2 + 15
\]

Solving for \( a \):

\[
a = -\frac{15}{400} = -\frac{3}{80}
\]

Thus, the equation of the ball's path is:

\[
y = -\frac{3}{80}x^2 + 15
\]

### Question 3: **Comparing the Paths**

To compare the two paths, we can observe the coefficients and vertex points:

- **Firework's Path:**  
  Equation: \( y = -\frac{1}{15}(x - 15)^2 + 15 \)  
  Vertex: \( (15, 15) \)

- **Ball's Path:**  
  Equation: \( y = -\frac{3}{80}x^2 + 15 \)  
  Vertex: \( (0, 15) \)

**Summary:**  
- The firework's path is symmetric around \( x = 15 \), and the ball's path is symmetric around \( x = 0 \).
- The firework reaches a maximum height of \( 15 \) meters at \( x = 15 \) meters, whereas the ball starts at the height of \( 15 \) meters and follows a downward trajectory.

Would you like to explore these equations further, or do you have any specific questions on these paths?

### Relative Questions:

1. How would the firework's path change if the launch site were not at the origin but at some height above the ground?
2. What would be the impact on the ball's path if it were thrown with an initial horizontal velocity?
3. Can you derive the time of flight for both the firework and the ball?
4. What is the significance of the coefficients \( a \) in both parabolic equations?
5. How does the horizontal distance of landing relate to the initial velocity of the projectile?

**Tip:** When analyzing parabolic paths, always identify the vertex and whether the parabola opens upwards or downwards, as this provides crucial insights into the trajectory.

Explore the parabolic paths of a firework and a ball launched from different points with specific heights and distances. Understand vertex forms of parabolas and projectile motion concepts. Suitable for high school students.

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