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The problem involves the trajectory of a basketball, and the given equation is the path of the basketball modeled by a parabola:

\[
y = \frac{-16x^2}{0.434v^2} + 1.15x + 8
\]
where:
- \( y \) is the height of the ball (in feet),
- \( x \) is the horizontal distance from the shooter (in feet),
- \( v \) is the initial velocity of the ball (in ft/sec),
- The basketball is shot from a height of 8 feet.

### Part (a): Find Initial Velocity \( v \)
The problem asks to find the initial velocity \( v \) if the basketball hoop is 10 feet high and located 16 feet away from the shooter. In this case:
- The height of the hoop \( y = 10 \) feet,
- The horizontal distance to the hoop \( x = 16 \) feet.

Plugging these values into the equation, we get:

\[
10 = \frac{-16(16)^2}{0.434v^2} + 1.15(16) + 8
\]

Now, we can solve this equation for \( v \).

### Part (b): Maximum Height
To find the maximum height, we will use the fact that the maximum height occurs at the vertex of the parabola. The x-coordinate of the vertex for a parabola of the form \( y = ax^2 + bx + c \) is given by:

\[
x_{\text{vertex}} = -\frac{b}{2a}
\]

We can use this to find the horizontal distance at which the maximum height occurs and then plug it back into the equation to find the maximum height.

Let me calculate these for you.The initial velocity \( v \) needed for the basketball to reach a 10-foot hoop located 16 feet away is approximately 23.99 ft/sec (taking the positive value, since velocity cannot be negative in this context).

### Part (b): Maximum Height

Next, let's find the maximum height of the basketball by determining the vertex of the parabola.The maximum height of the basketball is approximately 13.16 feet, and this occurs when the basketball is about 8.98 feet away from the shooter.

### Summary:
- **Initial velocity \( v \)**: Approximately 23.99 ft/sec.
- **Maximum height of the basketball**: 13.16 feet.

Let me know if you'd like further clarification or have any additional questions.

Here are 5 related questions to explore:
1. How would the result change if the shooter were farther from the hoop?
2. What happens to the trajectory if the initial velocity is increased?
3. How does the height from which the ball is shot (currently 8 feet) affect the maximum height?
4. Can the equation be adjusted for different gravitational constants (e.g., on the moon)?
5. What is the effect of air resistance on the trajectory?

**Tip:** When solving quadratic equations related to trajectories, remember that the vertex represents either the maximum or minimum point in the path, which is often important for physics problems involving projectile motion.

If a person shoots a basketball overhand from a position 8 feet above the floor, then the path of the basketball through the hoop can be modeled by the parabola y = (-16x^2 / 0.434v^2) + 1.15x + 8, where v is the initial velocity of the ball in ft/sec, y is the height of the ball and x is the distance away from the person. (a) If the basketball hoop is 10 ft high and located 16 ft away, what initial velocity v should the basketball have? (b) What is the maximum height of the basketball?

Solve a basketball trajectory problem where the path is modeled by a quadratic equation. Learn how to find the initial velocity needed for the basketball to reach the hoop and determine the maximum height. Suitable for high school students.

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