Question
What is the quadratic function that models the firework's trajectory?
Answers
Answered by
GPT-4o mini
To create a quadratic function that models the trajectory of a firework, we typically consider the standard form of a quadratic equation:
\[ y = ax^2 + bx + c \]
In the context of a firework's trajectory, the \( x \) variable represents the horizontal distance, and the \( y \) variable represents the height of the firework at that distance. The specific parameters \( a \), \( b \), and \( c \) will depend on the initial conditions of the firework launch, such as the initial height, the launch angle, and the initial velocity.
### Example:
1. **Setting Initial Conditions**:
- Suppose the firework is launched from the ground level, so the initial height (\( c \)) is 0.
- Assume it is launched at an angle and reaches a maximum height at a certain point.
2. **Constructing the function**:
- Let’s say the firework reaches its maximum height of 100 meters at a horizontal distance of 50 meters.
- We can also assume for simplicity that the firework lands back at the ground.
3. **Using vertex form**:
The vertex form of a quadratic equation is:
\[ y = a(x - h)^2 + k \]
where \( (h, k) \) is the vertex of the parabola. Here, \( h = 50 \) and \( k = 100 \).
4. To find the parameter \( a \):
Since it returns to the ground, we can make use of another point. The firework will be at ground level (\( y = 0 \)) when \( x = 0 \) and \( x = 100 \) (assuming it travels along a symmetrical path).
Using the point \( (0, 0) \):
\[ 0 = a(0 - 50)^2 + 100 \]
\[ 0 = 2500a + 100 \]
\[ 2500a = -100 \]
\[ a = -\frac{100}{2500} = -0.04 \]
5. **Final equation**:
Using the vertex form:
\[ y = -0.04(x - 50)^2 + 100 \]
This can also be expanded to standard form if necessary:
\[ y = -0.04(x^2 - 100x + 2500) + 100 \]
### Conclusion:
The quadratic function models the firework’s trajectory. Adjustments to the values of \( a \), \( b \), and \( c \) will change the shape and position of the trajectory, based on real initial conditions like speed, angle, and the height from which it is launched.
\[ y = ax^2 + bx + c \]
In the context of a firework's trajectory, the \( x \) variable represents the horizontal distance, and the \( y \) variable represents the height of the firework at that distance. The specific parameters \( a \), \( b \), and \( c \) will depend on the initial conditions of the firework launch, such as the initial height, the launch angle, and the initial velocity.
### Example:
1. **Setting Initial Conditions**:
- Suppose the firework is launched from the ground level, so the initial height (\( c \)) is 0.
- Assume it is launched at an angle and reaches a maximum height at a certain point.
2. **Constructing the function**:
- Let’s say the firework reaches its maximum height of 100 meters at a horizontal distance of 50 meters.
- We can also assume for simplicity that the firework lands back at the ground.
3. **Using vertex form**:
The vertex form of a quadratic equation is:
\[ y = a(x - h)^2 + k \]
where \( (h, k) \) is the vertex of the parabola. Here, \( h = 50 \) and \( k = 100 \).
4. To find the parameter \( a \):
Since it returns to the ground, we can make use of another point. The firework will be at ground level (\( y = 0 \)) when \( x = 0 \) and \( x = 100 \) (assuming it travels along a symmetrical path).
Using the point \( (0, 0) \):
\[ 0 = a(0 - 50)^2 + 100 \]
\[ 0 = 2500a + 100 \]
\[ 2500a = -100 \]
\[ a = -\frac{100}{2500} = -0.04 \]
5. **Final equation**:
Using the vertex form:
\[ y = -0.04(x - 50)^2 + 100 \]
This can also be expanded to standard form if necessary:
\[ y = -0.04(x^2 - 100x + 2500) + 100 \]
### Conclusion:
The quadratic function models the firework’s trajectory. Adjustments to the values of \( a \), \( b \), and \( c \) will change the shape and position of the trajectory, based on real initial conditions like speed, angle, and the height from which it is launched.