At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To determine the time it takes for the ball to hit the floor, we need to find the value of \( t \) when the height \( H(t) \) equals zero. The height can be modeled by the quadratic equation:
[tex]\[ H(t) = -16t^2 + 15t + 4 \][/tex]
We set \( H(t) = 0 \):
[tex]\[ -16t^2 + 15t + 4 = 0 \][/tex]
This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where:
- \( a = -16 \)
- \( b = 15 \)
- \( c = 4 \)
To solve for \( t \), we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, we compute the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
Substitute \( a = -16 \), \( b = 15 \), and \( c = 4 \) into the discriminant formula:
[tex]\[ \text{Discriminant} = 15^2 - 4(-16)(4) \][/tex]
[tex]\[ \text{Discriminant} = 225 + 256 \][/tex]
[tex]\[ \text{Sciminant} = 481 \][/tex]
Next, we substitute the values of \( a \), \( b \), and the discriminant back into the quadratic formula:
[tex]\[ t = \frac{-15 \pm \sqrt{481}}{2(-16)} \][/tex]
Calculating both potential solutions:
[tex]\[ t_1 = \frac{-15 + \sqrt{481}}{-32} \][/tex]
[tex]\[ t_2 = \frac{-15 - \sqrt{481}}{-32} \][/tex]
These yield the approximate solutions:
[tex]\[ t_1 \approx -0.217 \][/tex]
[tex]\[ t_2 \approx 1.154 \][/tex]
Given that time cannot be negative, we discard the negative solution. Therefore, the time it takes for the ball to hit the floor is approximately:
[tex]\[ t \approx 1.154 \][/tex]
So the closest answer to the options provided is:
1.15 seconds
[tex]\[ H(t) = -16t^2 + 15t + 4 \][/tex]
We set \( H(t) = 0 \):
[tex]\[ -16t^2 + 15t + 4 = 0 \][/tex]
This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where:
- \( a = -16 \)
- \( b = 15 \)
- \( c = 4 \)
To solve for \( t \), we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, we compute the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
Substitute \( a = -16 \), \( b = 15 \), and \( c = 4 \) into the discriminant formula:
[tex]\[ \text{Discriminant} = 15^2 - 4(-16)(4) \][/tex]
[tex]\[ \text{Discriminant} = 225 + 256 \][/tex]
[tex]\[ \text{Sciminant} = 481 \][/tex]
Next, we substitute the values of \( a \), \( b \), and the discriminant back into the quadratic formula:
[tex]\[ t = \frac{-15 \pm \sqrt{481}}{2(-16)} \][/tex]
Calculating both potential solutions:
[tex]\[ t_1 = \frac{-15 + \sqrt{481}}{-32} \][/tex]
[tex]\[ t_2 = \frac{-15 - \sqrt{481}}{-32} \][/tex]
These yield the approximate solutions:
[tex]\[ t_1 \approx -0.217 \][/tex]
[tex]\[ t_2 \approx 1.154 \][/tex]
Given that time cannot be negative, we discard the negative solution. Therefore, the time it takes for the ball to hit the floor is approximately:
[tex]\[ t \approx 1.154 \][/tex]
So the closest answer to the options provided is:
1.15 seconds
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.