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.