Question
A basketball is thrown with an initial upward velocity of 30
feet per second from a height of 6
feet above the ground. The equation h=-16t^2+30+6
models the height in feet t 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?
feet per second from a height of 6
feet above the ground. The equation h=-16t^2+30+6
models the height in feet t 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?
Answers
Answered by
GPT-4o mini
To solve for the time it takes for the basketball to reach a height of 10 feet after being thrown, we will start with the height equation provided:
\[
h = -16t^2 + 30t + 6
\]
We want to find the time \( t \) when the height \( h \) is 10 feet. Therefore, we set up the equation:
\[
10 = -16t^2 + 30t + 6
\]
Next, we can rearrange this equation to bring all terms to one side:
\[
-16t^2 + 30t + 6 - 10 = 0
\]
This simplifies to:
\[
-16t^2 + 30t - 4 = 0
\]
To make calculations easier, we can multiply the whole equation by -1:
\[
16t^2 - 30t + 4 = 0
\]
Now, we can use the quadratic formula to solve for \( t \):
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 16 \), \( b = -30 \), and \( c = 4 \). We start by calculating the discriminant:
\[
b^2 - 4ac = (-30)^2 - 4 \cdot 16 \cdot 4
\]
\[
= 900 - 256
\]
\[
= 644
\]
Now, we plug in the values into the quadratic formula:
\[
t = \frac{-(-30) \pm \sqrt{644}}{2 \cdot 16}
\]
\[
= \frac{30 \pm \sqrt{644}}{32}
\]
Next, we find \( \sqrt{644} \):
\[
\sqrt{644} = 2\sqrt{161}
\]
Then we can calculate \( t \):
\[
t = \frac{30 \pm 2\sqrt{161}}{32}
\]
To simplify further:
\[
t = \frac{30}{32} \pm \frac{2\sqrt{161}}{32}
\]
\[
= \frac{15}{16} \pm \frac{\sqrt{161}}{16}
\]
Thus, we have two potential solutions:
\[
t_1 = \frac{15 + \sqrt{161}}{16}
\]
\[
t_2 = \frac{15 - \sqrt{161}}{16}
\]
Now let's estimate \( \sqrt{161} \):
\[
\sqrt{161} \approx 12.6886 \quad (\text{since } 12.6886^2 \approx 161)
\]
Calculating \( t_1 \) and \( t_2 \):
1. For \( t_1 \):
\[
t_1 \approx \frac{15 + 12.6886}{16} \approx \frac{27.6886}{16} \approx 1.733 \text{ seconds}
\]
2. For \( t_2 \):
\[
t_2 \approx \frac{15 - 12.6886}{16} \approx \frac{2.3114}{16} \approx 0.144 \text{ seconds}
\]
Since the ball is thrown upward and then comes back down, we only consider the value of \( t_1 \). Therefore, the time it takes for the basketball to go into the hoop at a height of 10 feet is approximately:
\[
\boxed{1.73 \text{ seconds}}
\]
\[
h = -16t^2 + 30t + 6
\]
We want to find the time \( t \) when the height \( h \) is 10 feet. Therefore, we set up the equation:
\[
10 = -16t^2 + 30t + 6
\]
Next, we can rearrange this equation to bring all terms to one side:
\[
-16t^2 + 30t + 6 - 10 = 0
\]
This simplifies to:
\[
-16t^2 + 30t - 4 = 0
\]
To make calculations easier, we can multiply the whole equation by -1:
\[
16t^2 - 30t + 4 = 0
\]
Now, we can use the quadratic formula to solve for \( t \):
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 16 \), \( b = -30 \), and \( c = 4 \). We start by calculating the discriminant:
\[
b^2 - 4ac = (-30)^2 - 4 \cdot 16 \cdot 4
\]
\[
= 900 - 256
\]
\[
= 644
\]
Now, we plug in the values into the quadratic formula:
\[
t = \frac{-(-30) \pm \sqrt{644}}{2 \cdot 16}
\]
\[
= \frac{30 \pm \sqrt{644}}{32}
\]
Next, we find \( \sqrt{644} \):
\[
\sqrt{644} = 2\sqrt{161}
\]
Then we can calculate \( t \):
\[
t = \frac{30 \pm 2\sqrt{161}}{32}
\]
To simplify further:
\[
t = \frac{30}{32} \pm \frac{2\sqrt{161}}{32}
\]
\[
= \frac{15}{16} \pm \frac{\sqrt{161}}{16}
\]
Thus, we have two potential solutions:
\[
t_1 = \frac{15 + \sqrt{161}}{16}
\]
\[
t_2 = \frac{15 - \sqrt{161}}{16}
\]
Now let's estimate \( \sqrt{161} \):
\[
\sqrt{161} \approx 12.6886 \quad (\text{since } 12.6886^2 \approx 161)
\]
Calculating \( t_1 \) and \( t_2 \):
1. For \( t_1 \):
\[
t_1 \approx \frac{15 + 12.6886}{16} \approx \frac{27.6886}{16} \approx 1.733 \text{ seconds}
\]
2. For \( t_2 \):
\[
t_2 \approx \frac{15 - 12.6886}{16} \approx \frac{2.3114}{16} \approx 0.144 \text{ seconds}
\]
Since the ball is thrown upward and then comes back down, we only consider the value of \( t_1 \). Therefore, the time it takes for the basketball to go into the hoop at a height of 10 feet is approximately:
\[
\boxed{1.73 \text{ seconds}}
\]