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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 [tex]h = -16t^2 + 23t + 7[/tex] models the height in feet [tex]t[/tex] seconds after the basketball is thrown. After the ball passes its maximum height, it comes down and then goes into the hoop at a height of 10 feet above the ground. About how long after it was thrown does it go into the hoop?

A. 1.44 seconds
B. 0.15 seconds
C. 1.29 seconds
D. 1.70 seconds

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.

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Rewritten by : Jeany