DAREDEVIL DANNY is attempting a flaming hoop jump. Calculate the equation for the flaming hoop jump using the same method for the practice jump. Write the standard form of the equation that will get Daredevil Danny safely through the Flaming Hoop Jump of Awesome! a) Write the vertex form by substituting values for x, y, h, and k, and then solve for a. Show your work. b) Determine the values of a, b, and c. Show your work. c) Using your values of a, b, and c, did Daredevil Danny successfully pass through the Flaming Hoop Jump of Awesome? If not, describe what happened. Then, go back to critique your work and revise it. Write about your revision. d) Once you have determined the correct values of a, b, and c, verify algebraically that Daredevil Danny will successfully pass through the Flaming Hoop Jump of Awesome.
The Correct Answer and Explanation is :
To solve the problem of Daredevil Danny attempting the flaming hoop jump using a parabolic equation, let’s follow the steps outlined in your prompt. We will start by writing the equations, determining the coefficients, and finally concluding on Danny’s success.
Step A: Vertex Form
The vertex form of a parabola is given by:
[
y = a(x – h)^2 + k
]
Assuming the hoop’s center (vertex) is at ( (h, k) = (0, 0) ) for simplicity, and we want to find the value of ( a ). The equation for the jump can be influenced by the height ( y ) he needs to achieve to safely clear the hoop, which we’ll assume is at height ( k ). Let’s say the maximum height he needs to clear is 10 feet when he is at the center of the hoop.
So, substituting ( (0, 10) ) for ( (x, y) ):
[
10 = a(0 – 0)^2 + k
]
To find ( k ) at the vertex, we assume ( k = 10 ). Then we can plug in values:
[
10 = a(0)^2 + 10 \implies 10 = 10 \text{ (True)}
]
Now, let’s consider some values of ( x ) for points before and after the jump. If we choose ( x = 5 ) and expect ( y ) to be at least ( 0 ) (the height of the hoop), we substitute:
[
0 = a(5 – 0)^2 + 10
]
[
0 = 25a + 10
]
[
-10 = 25a \implies a = -\frac{2}{5}
]
Step B: Values of a, b, and c
The standard form of a parabola is:
[
y = ax^2 + bx + c
]
Using ( a = -\frac{2}{5} ):
- To convert from vertex to standard form, expand:
[
y = -\frac{2}{5}(x^2) + 10
]
This gives us:
[
y = -\frac{2}{5}x^2 + 10
]
Here, ( b = 0 ) and ( c = 10 ).
Thus, we have:
- ( a = -\frac{2}{5} )
- ( b = 0 )
- ( c = 10 )
Step C: Success of the Jump
Now, with ( a = -\frac{2}{5} ), ( b = 0 ), and ( c = 10 ), the equation is:
[
y = -\frac{2}{5}x^2 + 10
]
To find out if Danny successfully passed through the hoop, we evaluate the height ( y ) at the center (0 feet):
[
y = -\frac{2}{5}(0)^2 + 10 = 10
]
Since ( y = 10 ) is equal to the height of the hoop, Danny does indeed clear the hoop!
Step D: Verification
To verify that Daredevil Danny successfully clears the hoop, we can check for several values of ( x ) around the hoop:
- At ( x = 0 ):
[
y = -\frac{2}{5}(0)^2 + 10 = 10 \text{ (clears the hoop)}
] - At ( x = 5 ):
[
y = -\frac{2}{5}(5)^2 + 10 = -\frac{2}{5}(25) + 10 = -10 + 10 = 0 \text{ (just at the hoop)}
] - At ( x = 10 ):
[
y = -\frac{2}{5}(10)^2 + 10 = -\frac{2}{5}(100) + 10 = -40 + 10 = -30 \text{ (not successful)}
]
Conclusion
Daredevil Danny successfully clears the hoop at the center but hits the ground as he moves away from it. This shows the need for a better trajectory or adjustment to the jump’s starting speed or angle. Revising the approach to ensure a longer or higher arc may help achieve a safe landing.
In summary, we derived the parabolic equation, determined coefficients, assessed Danny’s performance, and confirmed the need for trajectory adjustments in his flaming hoop jump attempt.