Question 1180550: A programmer is writing the code for a new interactive basketball game. She is using quadratic relations to model the path of the ball. During the game, when a ball is shot, the path it follows is modelled by the quadratic relation, h = - 0.2d2 + 3d + 6, where h represented the height of the ball above the ground and d represented the distance of the ball from the shooter. Both distances are measured in feet.
a. How high was the ball when the shooter shot it?
b. What was the maximum height obtained by the ball?
c. A rim of a basketball net is 10 feet high. For what distance was the ball above the rim of the basketball net?
d. How far would the shooter have to be away from the rim of the basketball net for the ball to hit the rim of the basketball net and possibly go in?
e. Create a graphical model, using technology, to verify your calculations.
d. Create your own quadratic relation that would model the path of a shot from a distance of 15 feet that would hit the rim of the basketball net. Explain how you obtained your answer.
mathlover1 pls help
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's the solution to the basketball shot problem:
**a. Initial Height:**
When the shooter shoots the ball, the distance from the shooter (d) is 0. Substitute d = 0 into the equation:
h = -0.2(0)² + 3(0) + 6
h = 6 feet
The ball was 6 feet high when the shooter shot it.
**b. Maximum Height:**
The maximum height occurs at the vertex of the parabola. We can find the d-value of the vertex using the formula:
d = -b / 2a (where a = -0.2 and b = 3)
d = -3 / (2 * -0.2)
d = -3 / -0.4
d = 7.5 feet
Now, substitute d = 7.5 back into the height equation to find the maximum height:
h = -0.2(7.5)² + 3(7.5) + 6
h = -0.2(56.25) + 22.5 + 6
h = -11.25 + 22.5 + 6
h = 17.25 feet
The maximum height obtained by the ball was 17.25 feet.
**c. Distance above the rim:**
The rim is 10 feet high. We want to find the distances (d) where the ball is *above* 10 feet. So, we need to solve the inequality:
-0.2d² + 3d + 6 > 10
-0.2d² + 3d - 4 > 0
To find the boundary points, solve the quadratic equation:
-0.2d² + 3d - 4 = 0
Using the quadratic formula:
d = (-b ± √(b² - 4ac)) / 2a
d = (-3 ± √(3² - 4 * -0.2 * -4)) / (2 * -0.2)
d = (-3 ± √(9 - 3.2)) / -0.4
d = (-3 ± √5.8) / -0.4
d ≈ (-3 ± 2.41) / -0.4
This gives us two solutions for d:
d₁ ≈ (-3 + 2.41) / -0.4 ≈ 1.48 feet
d₂ ≈ (-3 - 2.41) / -0.4 ≈ 13.53 feet
The ball is above the rim between these two distances.
**d. Distance to hit the rim:**
We want to find the distances where the ball's height is exactly 10 feet:
-0.2d² + 3d + 6 = 10
-0.2d² + 3d - 4 = 0
We already solved this in part (c). The distances are approximately 1.48 feet and 13.53 feet.
**e. Graphical Model:**
You can use graphing software (like Desmos, GeoGebra, or even a graphing calculator) to plot the equation h = -0.2d² + 3d + 6. You can also plot the horizontal line h = 10 to visualize the rim. The intersection points of the parabola and the line will confirm the distances calculated above.
**f. Creating a Quadratic Relation:**
Here's how to create a quadratic relation for a shot from 15 feet that hits a 10-foot rim:
1. **Vertex Form:** A good starting point is the vertex form of a quadratic: h = a(d - k)² + j, where (k, j) is the vertex.
2. **Rim Condition:** Since the ball hits the rim at 15 feet (d = 15) and the rim is 10 feet high (h = 10), we know one point on the parabola.
3. **Shooter Condition:** We need to choose an initial height and adjust the equation accordingly. Let's assume the shooter releases the ball from a height of 6 feet. This gives us another point (0, 6).
4. **Solve for a, k, and j:** We have two points and three unknowns. Let's make an assumption to simplify the process. Assume the vertex of the parabola is somewhere between the shooter and the rim.
Let's assume the vertex is at d = 7.5. Since the shot starts at 6 feet, we know the vertex must be at a height greater than 6. Let's guess the vertex is at (7.5, 12).
Now we can use the vertex form and the point (15, 10):
10 = a(15 - 7.5)² + 12
-2 = a(7.5)²
a = -2 / 56.25 ≈ -0.0355
So, one possible relation is h = -0.0355(d - 7.5)² + 12. We can verify this by checking that h=6 when d=0.
There are many possible solutions since we had to make some assumptions. The key is to use the vertex form, incorporate the rim condition, and make a reasonable assumption about the vertex.
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