Question
A juggler tossed a ball upward into the air from a height of 4 feet at a velocity of 25 feet per second. After the ball started to come back down, the juggler caught it 5 feet above the ground.
To the nearest tenth of a second, how long was the ball in the air before the juggler caught it?
Hint: Use the formula h=
–
16t2+vt+s.
seconds
To the nearest tenth of a second, how long was the ball in the air before the juggler caught it?
Hint: Use the formula h=
–
16t2+vt+s.
seconds
Answers
GPT-4o mini
To determine how long the ball was in the air, we can use the kinematic equation provided:
\[
h(t) = -16t^2 + vt + s
\]
where:
- \( h(t) \) is the height of the ball at time \( t \),
- \( v \) is the initial velocity (25 feet/second),
- \( s \) is the initial height (4 feet),
- \( t \) is the time in seconds.
We need to find the time \( t \) when the ball reaches the height of 5 feet (when the juggler catches it). So, we set \( h(t) = 5 \):
\[
5 = -16t^2 + 25t + 4
\]
Rearranging this equation gives:
\[
-16t^2 + 25t + 4 - 5 = 0
\]
\[
-16t^2 + 25t - 1 = 0
\]
To solve this quadratic equation, we can apply the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \( a = -16 \), \( b = 25 \), and \( c = -1 \). Plugging in these values gives:
\[
t = \frac{-25 \pm \sqrt{25^2 - 4 \cdot (-16) \cdot (-1)}}{2 \cdot (-16)}
\]
Calculating the discriminant:
\[
\sqrt{625 - 64} = \sqrt{561}
\]
Finding the approximate value of \(\sqrt{561}\):
\[
\sqrt{561} \approx 23.7
\]
Now, substituting this back into the formula:
\[
t = \frac{-25 \pm 23.7}{-32}
\]
We will calculate the two potential solutions:
1. Using the positive root:
\[
t = \frac{-25 + 23.7}{-32} \approx \frac{-1.3}{-32} \approx 0.040625 \text{ (considered) }
\]
2. Using the negative root:
\[
t = \frac{-25 - 23.7}{-32} \approx \frac{-48.7}{-32} \approx 1.521875 \text{ seconds }
\]
The first solution \( t \approx 0.0406 \) seconds is likely the rising phase, while the second solution \( t \approx 1.5219 \) seconds is when the ball returns to the height of 5 feet.
Thus, the time the ball was in the air before being caught by the juggler is approximately:
\[
\boxed{1.5} \text{ seconds}
\]
\[
h(t) = -16t^2 + vt + s
\]
where:
- \( h(t) \) is the height of the ball at time \( t \),
- \( v \) is the initial velocity (25 feet/second),
- \( s \) is the initial height (4 feet),
- \( t \) is the time in seconds.
We need to find the time \( t \) when the ball reaches the height of 5 feet (when the juggler catches it). So, we set \( h(t) = 5 \):
\[
5 = -16t^2 + 25t + 4
\]
Rearranging this equation gives:
\[
-16t^2 + 25t + 4 - 5 = 0
\]
\[
-16t^2 + 25t - 1 = 0
\]
To solve this quadratic equation, we can apply the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \( a = -16 \), \( b = 25 \), and \( c = -1 \). Plugging in these values gives:
\[
t = \frac{-25 \pm \sqrt{25^2 - 4 \cdot (-16) \cdot (-1)}}{2 \cdot (-16)}
\]
Calculating the discriminant:
\[
\sqrt{625 - 64} = \sqrt{561}
\]
Finding the approximate value of \(\sqrt{561}\):
\[
\sqrt{561} \approx 23.7
\]
Now, substituting this back into the formula:
\[
t = \frac{-25 \pm 23.7}{-32}
\]
We will calculate the two potential solutions:
1. Using the positive root:
\[
t = \frac{-25 + 23.7}{-32} \approx \frac{-1.3}{-32} \approx 0.040625 \text{ (considered) }
\]
2. Using the negative root:
\[
t = \frac{-25 - 23.7}{-32} \approx \frac{-48.7}{-32} \approx 1.521875 \text{ seconds }
\]
The first solution \( t \approx 0.0406 \) seconds is likely the rising phase, while the second solution \( t \approx 1.5219 \) seconds is when the ball returns to the height of 5 feet.
Thus, the time the ball was in the air before being caught by the juggler is approximately:
\[
\boxed{1.5} \text{ seconds}
\]