A bb is shot downward from the top of a building leaving the barrel with an initial speed of 11.20 m/s downward. The top of the building is 68.4 m above the ground. How much time elapses between the instant it leaves the barrel and the instant of impact with the ground? Round your answer to 3 decimal places.
A bb is shot downward from the top of a building leaving the barrel with an initial speed of 11.20 m/s downward. The top of the building is 68.4 m above the ground. How much time elapses between the instant it leaves the barrel and the instant of impact with the ground? Round your answer to 3 decimal places.
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![## Physics Problem: Projectile Motion
**Problem Statement:**
A bb is shot downward from the top of a building leaving the barrel with an initial speed of 11.20 m/s downward. The top of the building is 68.4 m above the ground. How much time elapses between the instant it leaves the barrel and the instant of impact with the ground?
**Instructions:**
Round your answer to 3 decimal places.
---
**Solution:**
This problem involves solving for the time it takes for the bb to hit the ground using the equations of motion under constant acceleration, where gravity (g) is the acceleration. Let's break down the steps required to find the solution.
1. **Identify the known values:**
- Initial velocity \( v_0 = 11.20 \) m/s (downward)
- Initial height \( h = 68.4 \) m
- Acceleration due to gravity \( g = 9.81 \) m/s² (downward)
2. **Equation of motion:**
The equation to use is derived from the second equation of motion:
\[
h = v_0 t + \frac{1}{2} g t^2
\]
where \( h \) is the height, \( v_0 \) is the initial velocity, \( t \) is the time, and \( g \) is the acceleration due to gravity.
3. **Substitute the known values:**
\[
68.4 = 11.20 t + \frac{1}{2} \cdot 9.81 \cdot t^2
\]
4. **Rearrange the equation:**
\[
4.905t^2 + 11.20t - 68.4 = 0
\]
5. **Solve the quadratic equation:**
Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
t = \frac{-11.20 \pm \sqrt{(11.20)^2 - 4 \cdot 4.905 \cdot (-68.4)}}{2 \cdot 4.905}
\]
Calculate the discriminant:
\[
\Delta = (11.20)^2 - 4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f0c4e77-fecd-4d18-90ed-9b329310f28f%2F85c8aea7-3933-44b0-8cd3-4d979c14ab71%2Fprqte7d.png&w=3840&q=75)
Transcribed Image Text:## Physics Problem: Projectile Motion
**Problem Statement:**
A bb is shot downward from the top of a building leaving the barrel with an initial speed of 11.20 m/s downward. The top of the building is 68.4 m above the ground. How much time elapses between the instant it leaves the barrel and the instant of impact with the ground?
**Instructions:**
Round your answer to 3 decimal places.
---
**Solution:**
This problem involves solving for the time it takes for the bb to hit the ground using the equations of motion under constant acceleration, where gravity (g) is the acceleration. Let's break down the steps required to find the solution.
1. **Identify the known values:**
- Initial velocity \( v_0 = 11.20 \) m/s (downward)
- Initial height \( h = 68.4 \) m
- Acceleration due to gravity \( g = 9.81 \) m/s² (downward)
2. **Equation of motion:**
The equation to use is derived from the second equation of motion:
\[
h = v_0 t + \frac{1}{2} g t^2
\]
where \( h \) is the height, \( v_0 \) is the initial velocity, \( t \) is the time, and \( g \) is the acceleration due to gravity.
3. **Substitute the known values:**
\[
68.4 = 11.20 t + \frac{1}{2} \cdot 9.81 \cdot t^2
\]
4. **Rearrange the equation:**
\[
4.905t^2 + 11.20t - 68.4 = 0
\]
5. **Solve the quadratic equation:**
Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
t = \frac{-11.20 \pm \sqrt{(11.20)^2 - 4 \cdot 4.905 \cdot (-68.4)}}{2 \cdot 4.905}
\]
Calculate the discriminant:
\[
\Delta = (11.20)^2 - 4
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