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Answer :
To find out how long after it was thrown the basketball goes into the hoop at a height of 10 feet, we need to solve the equation for time, [tex]\( t \)[/tex], when the height [tex]\( h \)[/tex] is 10 feet.
The height equation given is:
[tex]\[ h(t) = -16t^2 + 23t + 7 \][/tex]
We want to determine when [tex]\( h(t) = 10 \)[/tex]:
[tex]\[ -16t^2 + 23t + 7 = 10 \][/tex]
Let's simplify this:
1. Subtract 10 from both sides to set the equation to zero:
[tex]\[ -16t^2 + 23t + 7 - 10 = 0 \][/tex]
[tex]\[ -16t^2 + 23t - 3 = 0 \][/tex]
Now, we need to solve this quadratic equation [tex]\(-16t^2 + 23t - 3 = 0\)[/tex].
To solve the quadratic equation, we can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 23 \)[/tex]
- [tex]\( c = -3 \)[/tex]
Calculating the discriminant:
[tex]\[ b^2 - 4ac = 23^2 - 4(-16)(-3) \][/tex]
[tex]\[ = 529 - 192 \][/tex]
[tex]\[ = 337 \][/tex]
Applying the quadratic formula:
[tex]\[ t = \frac{-23 \pm \sqrt{337}}{-32} \][/tex]
Solving for the positive root:
- One of the solutions will be positive, which makes sense in the context (since time cannot be negative).
The calculations give you the possible times:
- [tex]\( t \approx 0.15 \)[/tex] seconds
- [tex]\( t \approx 1.29 \)[/tex] seconds
Since we're interested in when the basketball first reaches the hoop height on the way down, the appropriate solution is:
[tex]\[ t \approx 1.29 \, \text{seconds} \][/tex]
Thus, the basketball goes into the hoop about 1.29 seconds after it was thrown.
The height equation given is:
[tex]\[ h(t) = -16t^2 + 23t + 7 \][/tex]
We want to determine when [tex]\( h(t) = 10 \)[/tex]:
[tex]\[ -16t^2 + 23t + 7 = 10 \][/tex]
Let's simplify this:
1. Subtract 10 from both sides to set the equation to zero:
[tex]\[ -16t^2 + 23t + 7 - 10 = 0 \][/tex]
[tex]\[ -16t^2 + 23t - 3 = 0 \][/tex]
Now, we need to solve this quadratic equation [tex]\(-16t^2 + 23t - 3 = 0\)[/tex].
To solve the quadratic equation, we can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 23 \)[/tex]
- [tex]\( c = -3 \)[/tex]
Calculating the discriminant:
[tex]\[ b^2 - 4ac = 23^2 - 4(-16)(-3) \][/tex]
[tex]\[ = 529 - 192 \][/tex]
[tex]\[ = 337 \][/tex]
Applying the quadratic formula:
[tex]\[ t = \frac{-23 \pm \sqrt{337}}{-32} \][/tex]
Solving for the positive root:
- One of the solutions will be positive, which makes sense in the context (since time cannot be negative).
The calculations give you the possible times:
- [tex]\( t \approx 0.15 \)[/tex] seconds
- [tex]\( t \approx 1.29 \)[/tex] seconds
Since we're interested in when the basketball first reaches the hoop height on the way down, the appropriate solution is:
[tex]\[ t \approx 1.29 \, \text{seconds} \][/tex]
Thus, the basketball goes into the hoop about 1.29 seconds after it was thrown.
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Rewritten by : Jeany
