A bb is shot downward from the top of a building leaving the barrel with an initial speed of 11.30 m/s downward. The top of the building is 60.7 m above the ground. How much time elapses between the instant it leaves the barrel and the instant of impact with the ground?
A bb is shot downward from the top of a building leaving the barrel with an initial speed of 11.30 m/s downward. The top of the building is 60.7 m above the ground. How much time elapses between the instant it leaves the barrel and the instant of impact with the ground?
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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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The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
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Question
![**Projectile Motion Problem**
**Problem Statement:**
A BB is shot downward from the top of a building, leaving the barrel with an initial speed of 11.30 m/s downward. The top of the building is 60.7 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.*
**Answer:**
To solve this problem, we use the equations of motion under uniform acceleration due to gravity. Using the given information, one can employ the following kinematic equation:
\[ y = V_{i} t + \frac{1}{2} a t^2 \]
Where:
- \( y \) is the displacement (60.7 m),
- \( V_{i} \) is the initial velocity (11.30 m/s downward),
- \( a \) is the acceleration due to gravity (9.8 m/s\(^2\)),
- \( t \) is the time which we need to find.
**Solution Steps:**
1. Substitute the given values into the kinematic equation:
\[ 60.7 = 11.3 t + \frac{1}{2} \times 9.8 t^2 \]
2. This simplifies to a quadratic equation in the form of \( at^2 + bt + c = 0 \):
\[ 4.9 t^2 + 11.3 t - 60.7 = 0 \]
3. Solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = 4.9 \)
- \( b = 11.3 \)
- \( c = -60.7 \)
4. Calculate the discriminant:
\[ \Delta = b^2 - 4ac \]
5. Determine the roots \( t \) by solving:
\[ t = \frac{-b \pm \sqrt{\Delta}}{2a} \]
After computing, round your final answer to 3 decimal places.
(Note: The actual calculations will yield the precise time of impact which students need to compute.)
**Graphs/Diagrams:**
Currently, there are no graphs or diagrams provided within this problem statement. However, a conceptual diagram may include a vertical line representing the height of the building with](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3b22dc5-b81c-452b-a8fe-ec73685697af%2F5c9243a1-6dab-4e2a-aafe-275616c98953%2Fed5d6lm.jpeg&w=3840&q=75)
Transcribed Image Text:**Projectile Motion Problem**
**Problem Statement:**
A BB is shot downward from the top of a building, leaving the barrel with an initial speed of 11.30 m/s downward. The top of the building is 60.7 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.*
**Answer:**
To solve this problem, we use the equations of motion under uniform acceleration due to gravity. Using the given information, one can employ the following kinematic equation:
\[ y = V_{i} t + \frac{1}{2} a t^2 \]
Where:
- \( y \) is the displacement (60.7 m),
- \( V_{i} \) is the initial velocity (11.30 m/s downward),
- \( a \) is the acceleration due to gravity (9.8 m/s\(^2\)),
- \( t \) is the time which we need to find.
**Solution Steps:**
1. Substitute the given values into the kinematic equation:
\[ 60.7 = 11.3 t + \frac{1}{2} \times 9.8 t^2 \]
2. This simplifies to a quadratic equation in the form of \( at^2 + bt + c = 0 \):
\[ 4.9 t^2 + 11.3 t - 60.7 = 0 \]
3. Solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = 4.9 \)
- \( b = 11.3 \)
- \( c = -60.7 \)
4. Calculate the discriminant:
\[ \Delta = b^2 - 4ac \]
5. Determine the roots \( t \) by solving:
\[ t = \frac{-b \pm \sqrt{\Delta}}{2a} \]
After computing, round your final answer to 3 decimal places.
(Note: The actual calculations will yield the precise time of impact which students need to compute.)
**Graphs/Diagrams:**
Currently, there are no graphs or diagrams provided within this problem statement. However, a conceptual diagram may include a vertical line representing the height of the building with
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