Hi, I need you to tell me what i have to do to find out the following parabolla. I have two points (0,1.63) and (4.572,3.05). This parabolla will be used to model a basketball shot, where the first set of coordinates i have given you is the coordinate from where it is thrown and the second set of coordinates is inside the basketball rim. Taking into account that the release angle is 52.2 degrees, find the equation for the parabolla step by step

To model the trajectory of the basketball shot with a parabolic equation, we need to determine the quadratic equation that passes through the given points and aligns with the given initial angle. A quadratic equation is generally of the form:

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

Given dataTo model the trajectory of the basketball shot with a parabolic equation, we need to determine the quadratic equation that passes through the given points and aligns with the given initial angle. A quadratic equation is generally of the form:

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

Given 
- Two points: \( (0, 1.63) \) (initial point of release) and \( (4.572, 3.05) \) (point at the rim).
- The release angle is \( 52.2^\circ \).

### Step-by-Step Process:

#### Step 1: Equation Setup
The quadratic equation for the parabola is \( y = ax^2 + bx + c \). Since the shot starts at point \( (0, 1.63) \), we can use this to immediately determine \( c \).

1. At \( x = 0 \), \( y = 1.63 \). Therefore, \( c = 1.63 \).

So, the equation becomes:

\[
y = ax^2 + bx + 1.63
\]

#### Step 2: Use the Second Point \( (4.572, 3.05) \)
Now, substitute \( x = 4.572 \) and \( y = 3.05 \) into the equation to find a relation between \( a \) and \( b \).

\[
3.05 = a(4.572)^2 + b(4.572) + 1.63
\]

Simplifying:

\[
3.05 = a(20.912) + 4.572b + 1.63
\]

\[
1.42 = 20.912a + 4.572b \tag{Equation 1}
\]

#### Step 3: Use the Initial Angle of 52.2°
The derivative \( \frac{dy}{dx} = 2ax + b \) gives the slope of the tangent to the parabola at any point. The slope of the tangent line at the release point \( x = 0 \) corresponds to the initial angle of \( 52.2^\circ \).

The slope \( m \) at the release point is \( \tan(52.2^\circ) \).

\[
\tan(52.2^\circ) \approx 1.292
\]

At \( x = 0 \), the slope is \( \frac{dy}{dx} = b \), so:

\[
b = 1.292 \tag{Equation 2}
\]

#### Step 4: Solve for \( a \)
Now substitute \( b = 1.292 \) into Equation 1:

\[
1.42 = 20.912a + 4.572(1.292)
\]

Simplifying:

\[
1.42 = 20.912a + 5.906
\]

\[
20.912a = 1.42 - 5.906
\]

\[
20.912a = -4.486
\]

\[
a = \frac{-4.486}{20.912} \approx -0.2146
\]

#### Step 5: Final Equation
Now that we have \( a \) and \( b \), we can write the final equation for the parabola:

\[
y = -0.2146x^2 + 1.292x + 1.63
\]

This is the equation that models the basketball shot.

---

### Next Steps and Validation
- The equation models the basketball trajectory based on the two points and the release angle. You can further validate it by checking if it predicts a realistic trajectory when plotted.

Would you like further details, a graph of this parabola, or have any questions?

### 5 Related Questions:
1. How can I find the maximum height of the basketball shot?
2. What would happen to the parabola if the release angle was lower?
3. How can I find the time it takes for the ball to reach the rim using this model?
4. What is the physical significance of the coefficients \( a \), \( b \), and \( c \) in this context?
5. Can we model air resistance in this quadratic equation?

### Tip:
When modeling projectile motion, the initial angle and velocity greatly influence the shape of the trajectory. Experimenting with different angles and velocities can help you understand the dynamics better.

Hi, I need you to tell me what I have to do to find out the following parabola. I have two points (0,1.63) and (4.572,3.05). This parabola will be used to model a basketball shot, where the first set of coordinates I have given you is the coordinate from where it is thrown, and the second set of coordinates is inside the basketball rim. Taking into account that the release angle is 52.2 degrees, find the equation for the parabola step by step.

Learn how to model the trajectory of a basketball shot with a quadratic equation. Using given points (0,1.63) and (4.572,3.05) and a release angle of 52.2°, we derive the equation step-by-step. Suitable for Grades 9-12 students studying projectile motion and quadratic equations.

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