Asked by Landon Sharp
Use the quadratic formula on the function h=−16t2+80t+3
ℎ
=
−
16
𝑡
2
+
80
𝑡
+
3
to calculate the time it will take for the baseball to hit the ground. Round to the nearest hundredth, if necessary.
Reference Sheet - Quadratic Formula
(1 point)
The baseball will hit the ground at about seconds.
ℎ
=
−
16
𝑡
2
+
80
𝑡
+
3
to calculate the time it will take for the baseball to hit the ground. Round to the nearest hundredth, if necessary.
Reference Sheet - Quadratic Formula
(1 point)
The baseball will hit the ground at about seconds.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the time \( t \) when the baseball hits the ground, we need to set the height \( h \) to zero and solve the quadratic equation:
\[
-16t^2 + 80t + 3 = 0
\]
We can use the quadratic formula, which is given by:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In the equation, \( a = -16 \), \( b = 80 \), and \( c = 3 \). Let's calculate the discriminant \( b^2 - 4ac \):
1. Calculate \( b^2 \):
\[
b^2 = 80^2 = 6400
\]
2. Calculate \( 4ac \):
\[
4ac = 4 \cdot (-16) \cdot 3 = -192
\]
3. Now calculate the discriminant:
\[
b^2 - 4ac = 6400 - (-192) = 6400 + 192 = 6592
\]
Now we can use the quadratic formula:
\[
t = \frac{-80 \pm \sqrt{6592}}{2 \cdot (-16)}
\]
Let's compute \( \sqrt{6592} \):
\[
\sqrt{6592} \approx 81.1 \quad \text{(approximately)}
\]
Now, substituting back into the formula:
\[
t = \frac{-80 \pm 81.1}{-32}
\]
This gives us two potential solutions:
1. Using the plus sign:
\[
t_1 = \frac{-80 + 81.1}{-32} = \frac{1.1}{-32} \approx -0.034375 \quad \text{(not a valid solution, as time cannot be negative)}
\]
2. Using the minus sign:
\[
t_2 = \frac{-80 - 81.1}{-32} = \frac{-161.1}{-32} \approx 5.03
\]
So rounding to the nearest hundredth:
\[
t \approx 5.03 \text{ seconds}
\]
Therefore, the baseball will hit the ground at about **5.03 seconds**.
\[
-16t^2 + 80t + 3 = 0
\]
We can use the quadratic formula, which is given by:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In the equation, \( a = -16 \), \( b = 80 \), and \( c = 3 \). Let's calculate the discriminant \( b^2 - 4ac \):
1. Calculate \( b^2 \):
\[
b^2 = 80^2 = 6400
\]
2. Calculate \( 4ac \):
\[
4ac = 4 \cdot (-16) \cdot 3 = -192
\]
3. Now calculate the discriminant:
\[
b^2 - 4ac = 6400 - (-192) = 6400 + 192 = 6592
\]
Now we can use the quadratic formula:
\[
t = \frac{-80 \pm \sqrt{6592}}{2 \cdot (-16)}
\]
Let's compute \( \sqrt{6592} \):
\[
\sqrt{6592} \approx 81.1 \quad \text{(approximately)}
\]
Now, substituting back into the formula:
\[
t = \frac{-80 \pm 81.1}{-32}
\]
This gives us two potential solutions:
1. Using the plus sign:
\[
t_1 = \frac{-80 + 81.1}{-32} = \frac{1.1}{-32} \approx -0.034375 \quad \text{(not a valid solution, as time cannot be negative)}
\]
2. Using the minus sign:
\[
t_2 = \frac{-80 - 81.1}{-32} = \frac{-161.1}{-32} \approx 5.03
\]
So rounding to the nearest hundredth:
\[
t \approx 5.03 \text{ seconds}
\]
Therefore, the baseball will hit the ground at about **5.03 seconds**.
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