Question
The CHS Baseball team was on the field and the batter popped the ball up. The equation b(t)=80t-16t^2+3.5
represents the height of the ball above the ground in feet as a function of time in seconds. How long will the catcher have to get in position to catch the ball before it hits the ground? Round to the nearest second.
represents the height of the ball above the ground in feet as a function of time in seconds. How long will the catcher have to get in position to catch the ball before it hits the ground? Round to the nearest second.
Answers
GPT-4o mini
To determine how long the catcher has to get in position to catch the ball before it hits the ground, we need to find the time \( t \) when the height \( b(t) \) is equal to 0. The equation provided is:
\[
b(t) = 80t - 16t^2 + 3.5
\]
To find when the ball hits the ground, we set \( b(t) \) equal to zero and solve for \( t \):
\[
80t - 16t^2 + 3.5 = 0
\]
We can rearrange this into standard quadratic form:
\[
-16t^2 + 80t + 3.5 = 0
\]
For easier computation, we can multiply through by -1:
\[
16t^2 - 80t - 3.5 = 0
\]
Now, we can use the quadratic formula to solve for \( t \):
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here:
- \( a = 16 \)
- \( b = -80 \)
- \( c = -3.5 \)
Calculating the discriminant \( b^2 - 4ac \):
\[
b^2 = (-80)^2 = 6400
\]
\[
4ac = 4 \cdot 16 \cdot (-3.5) = -224
\]
\[
b^2 - 4ac = 6400 + 224 = 6624
\]
Now we can substitute back into the quadratic formula:
\[
t = \frac{-(-80) \pm \sqrt{6624}}{2 \cdot 16}
\]
\[
t = \frac{80 \pm \sqrt{6624}}{32}
\]
Calculating \( \sqrt{6624} \):
\[
\sqrt{6624} \approx 81.5
\]
Now substituting back:
\[
t \approx \frac{80 \pm 81.5}{32}
\]
Calculating the two potential solutions:
1. \( t_1 = \frac{80 + 81.5}{32} = \frac{161.5}{32} \approx 5.04 \)
2. \( t_2 = \frac{80 - 81.5}{32} = \frac{-1.5}{32} \approx -0.05 \) (This solution is not physically meaningful since time cannot be negative).
Thus, the only valid solution is approximately \( t \approx 5.04 \) seconds. Rounding to the nearest second, the catcher will have to get in position to catch the ball in approximately:
\[
\boxed{5} \text{ seconds}
\]
\[
b(t) = 80t - 16t^2 + 3.5
\]
To find when the ball hits the ground, we set \( b(t) \) equal to zero and solve for \( t \):
\[
80t - 16t^2 + 3.5 = 0
\]
We can rearrange this into standard quadratic form:
\[
-16t^2 + 80t + 3.5 = 0
\]
For easier computation, we can multiply through by -1:
\[
16t^2 - 80t - 3.5 = 0
\]
Now, we can use the quadratic formula to solve for \( t \):
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here:
- \( a = 16 \)
- \( b = -80 \)
- \( c = -3.5 \)
Calculating the discriminant \( b^2 - 4ac \):
\[
b^2 = (-80)^2 = 6400
\]
\[
4ac = 4 \cdot 16 \cdot (-3.5) = -224
\]
\[
b^2 - 4ac = 6400 + 224 = 6624
\]
Now we can substitute back into the quadratic formula:
\[
t = \frac{-(-80) \pm \sqrt{6624}}{2 \cdot 16}
\]
\[
t = \frac{80 \pm \sqrt{6624}}{32}
\]
Calculating \( \sqrt{6624} \):
\[
\sqrt{6624} \approx 81.5
\]
Now substituting back:
\[
t \approx \frac{80 \pm 81.5}{32}
\]
Calculating the two potential solutions:
1. \( t_1 = \frac{80 + 81.5}{32} = \frac{161.5}{32} \approx 5.04 \)
2. \( t_2 = \frac{80 - 81.5}{32} = \frac{-1.5}{32} \approx -0.05 \) (This solution is not physically meaningful since time cannot be negative).
Thus, the only valid solution is approximately \( t \approx 5.04 \) seconds. Rounding to the nearest second, the catcher will have to get in position to catch the ball in approximately:
\[
\boxed{5} \text{ seconds}
\]