The Table Shows The Height, In Meters, Of An Object That Is Dropped As Time Passes Until The Object Hits The Ground.Falling Object$\[ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \begin{tabular}{c} Time (sec) \\ $x$ \end{tabular} & 0 & 0.5 &
Introduction
In this article, we will be analyzing a table that shows the height, in meters, of an object that is dropped as time passes until the object hits the ground. The table provides us with a set of data that we can use to calculate the height of the object at different times. We will be using mathematical concepts and formulas to analyze the data and draw conclusions about the object's motion.
The Table
| Time (sec) | Height (m) |
|---|---|
| 0 | 10 |
| 0.5 | 8.5 |
| 1 | 7 |
| 1.5 | 5.5 |
| 2 | 4 |
| 2.5 | 2.5 |
| 3 | 1 |
| 3.5 | 0.5 |
| 4 | 0 |
Understanding the Data
The table shows the height of the object at different times. The time is measured in seconds, and the height is measured in meters. We can see that the height of the object decreases as time passes. This is because the object is accelerating downward due to gravity.
Calculating the Height
To calculate the height of the object at a given time, we can use the formula:
h(t) = h0 - gt^2
where h(t) is the height at time t, h0 is the initial height, g is the acceleration due to gravity, and t is the time.
We can use this formula to calculate the height of the object at each time in the table.
Calculating the Acceleration
To calculate the acceleration of the object, we can use the formula:
a = Δv / Δt
where a is the acceleration, Δv is the change in velocity, and Δt is the change in time.
We can use this formula to calculate the acceleration of the object at each time in the table.
Analyzing the Data
Let's analyze the data in the table.
| Time (sec) | Height (m) | Δh | Δt | a |
|---|---|---|---|---|
| 0 | 10 | - | - | - |
| 0.5 | 8.5 | -1.5 | 0.5 | -3 |
| 1 | 7 | -1.5 | 0.5 | -3 |
| 1.5 | 5.5 | -1.5 | 0.5 | -3 |
| 2 | 4 | -1.5 | 0.5 | -3 |
| 2.5 | 2.5 | -1.5 | 0.5 | -3 |
| 3 | 1 | -1.5 | 0.5 | -3 |
| 3.5 | 0.5 | -0.5 | 0.5 | -1 |
| 4 | 0 | -0.5 | 0.5 | -1 |
We can see that the acceleration of the object is constant at -3 m/s^2 for the first 2 seconds, and then it decreases to -1 m/s^2 for the last 2 seconds.
Conclusion
In this article, we analyzed a table that shows the height of an object that is dropped as time passes until the object hits the ground. We used mathematical concepts and formulas to calculate the height and acceleration of the object at different times. We found that the acceleration of the object is constant at -3 m/s^2 for the first 2 seconds, and then it decreases to -1 m/s^2 for the last 2 seconds.
References
- [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of physics. John Wiley & Sons.
- [2] Serway, R. A., & Jewett, J. W. (2018). Physics for scientists and engineers. Cengage Learning.
Discussion
This problem is a classic example of an object under constant acceleration. The acceleration due to gravity is a constant 9.8 m/s^2, and the object's velocity increases by 9.8 m/s every second. The object's height decreases by 4.9 m every second, which is half of the velocity.
The table shows that the object's height decreases by 1.5 m every 0.5 seconds, which is consistent with the acceleration due to gravity. The acceleration of the object is constant at -3 m/s^2 for the first 2 seconds, and then it decreases to -1 m/s^2 for the last 2 seconds.
This problem can be solved using the equation of motion:
s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration.
We can use this equation to calculate the displacement of the object at each time in the table.
Solution
Let's solve the problem using the equation of motion.
| Time (sec) | Height (m) | Δh | Δt | a |
|---|---|---|---|---|
| 0 | 10 | - | - | - |
| 0.5 | 8.5 | -1.5 | 0.5 | -3 |
| 1 | 7 | -1.5 | 0.5 | -3 |
| 1.5 | 5.5 | -1.5 | 0.5 | -3 |
| 2 | 4 | -1.5 | 0.5 | -3 |
| 2.5 | 2.5 | -1.5 | 0.5 | -3 |
| 3 | 1 | -1.5 | 0.5 | -3 |
| 3.5 | 0.5 | -0.5 | 0.5 | -1 |
| 4 | 0 | -0.5 | 0.5 | -1 |
We can see that the displacement of the object at each time is consistent with the acceleration due to gravity.
Conclusion
Q: What is the initial height of the object?
A: The initial height of the object is 10 meters.
Q: What is the acceleration due to gravity?
A: The acceleration due to gravity is 9.8 m/s^2.
Q: How does the height of the object change over time?
A: The height of the object decreases over time. At each time interval, the height decreases by 1.5 meters.
Q: What is the acceleration of the object at each time interval?
A: The acceleration of the object is constant at -3 m/s^2 for the first 2 seconds, and then it decreases to -1 m/s^2 for the last 2 seconds.
Q: How can we calculate the height of the object at each time?
A: We can use the formula:
h(t) = h0 - gt^2
where h(t) is the height at time t, h0 is the initial height, g is the acceleration due to gravity, and t is the time.
Q: What is the equation of motion for an object under constant acceleration?
A: The equation of motion is:
s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration.
Q: How can we use the equation of motion to calculate the displacement of the object at each time?
A: We can plug in the values of u, t, and a into the equation of motion to calculate the displacement of the object at each time.
Q: What is the significance of the acceleration due to gravity?
A: The acceleration due to gravity is a fundamental constant that describes the force of gravity on an object. It is a key concept in physics and is used to describe the motion of objects under the influence of gravity.
Q: How does the acceleration of the object change over time?
A: The acceleration of the object is constant at -3 m/s^2 for the first 2 seconds, and then it decreases to -1 m/s^2 for the last 2 seconds.
Q: What is the relationship between the height of the object and the time?
A: The height of the object decreases over time. At each time interval, the height decreases by 1.5 meters.
Q: How can we use the data in the table to calculate the acceleration of the object?
A: We can use the formula:
a = Δv / Δt
where a is the acceleration, Δv is the change in velocity, and Δt is the change in time.
Q: What is the significance of the equation of motion?
A: The equation of motion is a fundamental concept in physics that describes the motion of objects under the influence of forces. It is used to calculate the displacement, velocity, and acceleration of an object over time.
Q: How can we use the equation of motion to solve problems involving objects under constant acceleration?
A: We can plug in the values of u, t, and a into the equation of motion to calculate the displacement of the object at each time.
Q: What is the relationship between the acceleration of the object and the time?
A: The acceleration of the object is constant at -3 m/s^2 for the first 2 seconds, and then it decreases to -1 m/s^2 for the last 2 seconds.
Q: How can we use the data in the table to calculate the height of the object at each time?
A: We can use the formula:
h(t) = h0 - gt^2
where h(t) is the height at time t, h0 is the initial height, g is the acceleration due to gravity, and t is the time.