Thank you for visiting A basketball is thrown with an initial upward velocity of 23 feet per second from a height of 7 feet above the ground The equation. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
To find out how long it takes for the basketball to go into the hoop at a height of 10 feet above the ground, we need to use the given equation for the height of the ball:
[tex]\[ h = -16t^2 + 23t + 7 \][/tex]
The ball needs to reach a height of 10 feet, so we set [tex]\( h \)[/tex] to 10 and solve for [tex]\( t \)[/tex]:
[tex]\[ 10 = -16t^2 + 23t + 7 \][/tex]
Rearrange the equation to form a standard quadratic equation:
[tex]\[ 0 = -16t^2 + 23t + 7 - 10 \][/tex]
[tex]\[ 0 = -16t^2 + 23t - 3 \][/tex]
Now, let's solve this quadratic equation using the quadratic formula which is:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this equation, the coefficients are:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 23 \)[/tex]
- [tex]\( c = -3 \)[/tex]
First, calculate the discriminant ([tex]\( b^2 - 4ac \)[/tex]):
[tex]\[ b^2 = 23^2 = 529 \][/tex]
[tex]\[ 4ac = 4 \times (-16) \times (-3) = 192 \][/tex]
[tex]\[ b^2 - 4ac = 529 - 192 = 337 \][/tex]
Since the discriminant is positive, the quadratic equation has two real solutions. Now we calculate the two possible values for [tex]\( t \)[/tex] (time):
[tex]\[ t = \frac{-23 \pm \sqrt{337}}{2 \times -16} \][/tex]
Calculate the two roots:
1. [tex]\( t_1 = \frac{-23 + \sqrt{337}}{-32} \approx 1.29 \)[/tex] seconds
2. [tex]\( t_2 = \frac{-23 - \sqrt{337}}{-32} \approx 0.15 \)[/tex] seconds
Since time cannot be negative and we're interested in the time after the ball is thrown and passes its maximum height, we choose the positive value that makes sense in this context:
The ball goes into the hoop about [tex]\( 1.29 \)[/tex] seconds after it was thrown.
[tex]\[ h = -16t^2 + 23t + 7 \][/tex]
The ball needs to reach a height of 10 feet, so we set [tex]\( h \)[/tex] to 10 and solve for [tex]\( t \)[/tex]:
[tex]\[ 10 = -16t^2 + 23t + 7 \][/tex]
Rearrange the equation to form a standard quadratic equation:
[tex]\[ 0 = -16t^2 + 23t + 7 - 10 \][/tex]
[tex]\[ 0 = -16t^2 + 23t - 3 \][/tex]
Now, let's solve this quadratic equation using the quadratic formula which is:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this equation, the coefficients are:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 23 \)[/tex]
- [tex]\( c = -3 \)[/tex]
First, calculate the discriminant ([tex]\( b^2 - 4ac \)[/tex]):
[tex]\[ b^2 = 23^2 = 529 \][/tex]
[tex]\[ 4ac = 4 \times (-16) \times (-3) = 192 \][/tex]
[tex]\[ b^2 - 4ac = 529 - 192 = 337 \][/tex]
Since the discriminant is positive, the quadratic equation has two real solutions. Now we calculate the two possible values for [tex]\( t \)[/tex] (time):
[tex]\[ t = \frac{-23 \pm \sqrt{337}}{2 \times -16} \][/tex]
Calculate the two roots:
1. [tex]\( t_1 = \frac{-23 + \sqrt{337}}{-32} \approx 1.29 \)[/tex] seconds
2. [tex]\( t_2 = \frac{-23 - \sqrt{337}}{-32} \approx 0.15 \)[/tex] seconds
Since time cannot be negative and we're interested in the time after the ball is thrown and passes its maximum height, we choose the positive value that makes sense in this context:
The ball goes into the hoop about [tex]\( 1.29 \)[/tex] seconds after it was thrown.
Thank you for reading the article A basketball is thrown with an initial upward velocity of 23 feet per second from a height of 7 feet above the ground The equation. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
- You are operating a recreational vessel less than 39 4 feet long on federally controlled waters Which of the following is a legal sound device
- Which step should a food worker complete to prevent cross contact when preparing and serving an allergen free meal A Clean and sanitize all surfaces
- For one month Siera calculated her hometown s average high temperature in degrees Fahrenheit She wants to convert that temperature from degrees Fahrenheit to degrees
Rewritten by : Jeany
