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Answer :
To solve this problem, we need to determine the time [tex]\( t \)[/tex] at which the basketball reaches a height of 10 feet after being thrown. The height of the basketball as a function of time is given by the equation:
[tex]\[ h = -16t^2 + 23t + 7 \][/tex]
We are asked to find the time [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 10 feet. So, we set [tex]\( h \)[/tex] to 10 and solve the equation:
[tex]\[ 10 = -16t^2 + 23t + 7 \][/tex]
First, we need to rearrange this into standard quadratic form:
[tex]\[ -16t^2 + 23t + 7 - 10 = 0 \][/tex]
[tex]\[ -16t^2 + 23t - 3 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 23 \)[/tex], and [tex]\( c = -3 \)[/tex].
To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ t = \frac{-23 \pm \sqrt{23^2 - 4(-16)(-3)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-23 \pm \sqrt{529 - 192}}{-32} \][/tex]
[tex]\[ t = \frac{-23 \pm \sqrt{337}}{-32} \][/tex]
Solving inside the square root:
[tex]\[ \sqrt{337} \approx 18.36 \][/tex]
Now, we have two possible solutions for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{-23 + 18.36}{-32} \approx \frac{-4.64}{-32} \approx 0.145 \][/tex]
[tex]\[ t_2 = \frac{-23 - 18.36}{-32} \approx \frac{-41.36}{-32} \approx 1.292 \][/tex]
Since the basketball is thrown upwards and then falls down, there are two times when it reaches the height of 10 feet. The basketball first reaches a height of 10 feet at approximately 0.145 seconds after it is thrown and again at approximately 1.292 seconds after it is thrown.
Since we're looking for the time after the ball passes its maximum height and then comes down into the hoop, we consider the second time:
[tex]\[ t \approx 1.292 \text{ seconds} \][/tex]
Therefore, the basketball goes into the hoop at approximately [tex]\(1.29\)[/tex] seconds after it was thrown. The correct answer is:
[tex]\[ 1.29 \text{ seconds} \][/tex]
[tex]\[ h = -16t^2 + 23t + 7 \][/tex]
We are asked to find the time [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 10 feet. So, we set [tex]\( h \)[/tex] to 10 and solve the equation:
[tex]\[ 10 = -16t^2 + 23t + 7 \][/tex]
First, we need to rearrange this into standard quadratic form:
[tex]\[ -16t^2 + 23t + 7 - 10 = 0 \][/tex]
[tex]\[ -16t^2 + 23t - 3 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 23 \)[/tex], and [tex]\( c = -3 \)[/tex].
To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ t = \frac{-23 \pm \sqrt{23^2 - 4(-16)(-3)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-23 \pm \sqrt{529 - 192}}{-32} \][/tex]
[tex]\[ t = \frac{-23 \pm \sqrt{337}}{-32} \][/tex]
Solving inside the square root:
[tex]\[ \sqrt{337} \approx 18.36 \][/tex]
Now, we have two possible solutions for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{-23 + 18.36}{-32} \approx \frac{-4.64}{-32} \approx 0.145 \][/tex]
[tex]\[ t_2 = \frac{-23 - 18.36}{-32} \approx \frac{-41.36}{-32} \approx 1.292 \][/tex]
Since the basketball is thrown upwards and then falls down, there are two times when it reaches the height of 10 feet. The basketball first reaches a height of 10 feet at approximately 0.145 seconds after it is thrown and again at approximately 1.292 seconds after it is thrown.
Since we're looking for the time after the ball passes its maximum height and then comes down into the hoop, we consider the second time:
[tex]\[ t \approx 1.292 \text{ seconds} \][/tex]
Therefore, the basketball goes into the hoop at approximately [tex]\(1.29\)[/tex] seconds after it was thrown. The correct answer is:
[tex]\[ 1.29 \text{ seconds} \][/tex]
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!
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