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Answer :
To solve the problem of finding out how long it takes for the basketball to go into the hoop at a height of 10 feet after being thrown, we need to use the given height equation:
[tex]\[ h = -16t^2 + 23t + 7 \][/tex]
This equation gives the height [tex]\( h \)[/tex] of the basketball [tex]\( t \)[/tex] seconds after it is thrown. We are looking for the time [tex]\( t \)[/tex] when the basketball's height is 10 feet, hence we set [tex]\( h \)[/tex] to 10 and solve the equation:
[tex]\[ -16t^2 + 23t + 7 = 10 \][/tex]
Rearrange the equation for a standard quadratic form:
[tex]\[ -16t^2 + 23t + 7 - 10 = 0 \][/tex]
Which simplifies to:
[tex]\[ -16t^2 + 23t - 3 = 0 \][/tex]
Now, we solve this quadratic equation using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where [tex]\( a = -16 \)[/tex], [tex]\( b = 23 \)[/tex], and [tex]\( c = -3 \)[/tex].
First, we find the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = 23^2 - 4 \times (-16) \times (-3) \][/tex]
Calculate the square of [tex]\( b \)[/tex]:
[tex]\[ 23^2 = 529 \][/tex]
Calculate [tex]\( 4ac \)[/tex]:
[tex]\[ 4 \times 16 \times 3 = 192 \][/tex]
Subtract:
[tex]\[ 529 - 192 = 337 \][/tex]
Now, with the discriminant calculated, we can find the roots using the quadratic formula:
Calculate the square root of the discriminant:
[tex]\[ \sqrt{337} \][/tex]
Then solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-23 \pm \sqrt{337}}{-32} \][/tex]
There are two possible solutions for [tex]\( t \)[/tex], one positive and one negative. Since time cannot be negative, we only consider the positive solution:
The positive solution gives:
[tex]\[ t \approx 1.29 \text{ seconds} \][/tex]
Therefore, the basketball goes into the hoop approximately 1.29 seconds after it is thrown.
[tex]\[ h = -16t^2 + 23t + 7 \][/tex]
This equation gives the height [tex]\( h \)[/tex] of the basketball [tex]\( t \)[/tex] seconds after it is thrown. We are looking for the time [tex]\( t \)[/tex] when the basketball's height is 10 feet, hence we set [tex]\( h \)[/tex] to 10 and solve the equation:
[tex]\[ -16t^2 + 23t + 7 = 10 \][/tex]
Rearrange the equation for a standard quadratic form:
[tex]\[ -16t^2 + 23t + 7 - 10 = 0 \][/tex]
Which simplifies to:
[tex]\[ -16t^2 + 23t - 3 = 0 \][/tex]
Now, we solve this quadratic equation using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where [tex]\( a = -16 \)[/tex], [tex]\( b = 23 \)[/tex], and [tex]\( c = -3 \)[/tex].
First, we find the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = 23^2 - 4 \times (-16) \times (-3) \][/tex]
Calculate the square of [tex]\( b \)[/tex]:
[tex]\[ 23^2 = 529 \][/tex]
Calculate [tex]\( 4ac \)[/tex]:
[tex]\[ 4 \times 16 \times 3 = 192 \][/tex]
Subtract:
[tex]\[ 529 - 192 = 337 \][/tex]
Now, with the discriminant calculated, we can find the roots using the quadratic formula:
Calculate the square root of the discriminant:
[tex]\[ \sqrt{337} \][/tex]
Then solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-23 \pm \sqrt{337}}{-32} \][/tex]
There are two possible solutions for [tex]\( t \)[/tex], one positive and one negative. Since time cannot be negative, we only consider the positive solution:
The positive solution gives:
[tex]\[ t \approx 1.29 \text{ seconds} \][/tex]
Therefore, the basketball goes into the hoop approximately 1.29 seconds after it is thrown.
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Rewritten by : Jeany