Question
A small object is dropped from the top of a 200-foot cliff. A measuring device on the object records the height of the object above the ground every second. The following table shows the results for the first 6 seconds.
Time (Seconds) 0 1 2 3 4 5 6
Height (Meters) 200.0 195.1 180.4 155.9 121.6 77.5 23.6
A. Model this scenario with a function. Explain how you chose which function to use.
Time (Seconds) 0 1 2 3 4 5 6
Height (Meters) 200.0 195.1 180.4 155.9 121.6 77.5 23.6
A. Model this scenario with a function. Explain how you chose which function to use.
Answers
GPT-4o mini
To model the height of an object dropped from a height, we can start by analyzing the given data and identifying the appropriate function to represent the height as a function of time.
### Data Overview
Here are the height measurements converted from feet to meters:
- The object starts at 200 feet, which is approximately 60.96 meters.
- The height decreases as time progresses, indicating a downward motion due to gravity.
### Choosing a Function
In free fall (neglecting air resistance), the height \( h(t) \) of an object at time \( t \) is generally modeled using the equations of motion under constant acceleration. The equation can be expressed as:
\[
h(t) = h_0 - \frac{1}{2} g t^2
\]
where:
- \( h_0 \) is the initial height,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) or about \( 10 \, \text{m/s}^2 \) for simplicity),
- \( t \) is the time in seconds.
### Conversion and Simplification
Given that the initial height \( h_0 \) is approximately \( 60.96 \) meters, we can represent the height function based on the data and we'll need to find a suitable value for \( g \).
### Fitting the Function
Using the data provided, let's initially establish whether a quadratic model is appropriate by observing the heights. The heights are decreasing in a somewhat quadratic manner, meaning that after 6 seconds, the decrease isn’t linear, which aligns well with a quadratic function.
We can refine the constant for gravity if we observe the given heights against the parameterized equation \( h(t) = 60.96 - \frac{1}{2} g t^2 \).
### Example Estimation
1. **From the Data:**
- At \( t = 0 \): \( h(0) = 200 \)
- At \( t = 1 \): \( h(1) \approx 195.1 \)
- At \( t = 2 \): \( h(2) \approx 180.4 \)
- At \( t = 3 \): \( h(3) \approx 155.9 \)
- At \( t = 4 \): \( h(4) \approx 121.6 \)
- At \( t = 5 \): \( h(5) \approx 77.5 \)
- At \( t = 6 \): \( h(6) \approx 23.6 \)
### Formulating the Function
To fit our data, we can perform a polynomial regression if necessary or simply derive constants through substitution and solving:
Assuming \( g = 9.81 \, \text{m/s}^2\) (or an average approximation similar), we can fit our model accordingly incorporating our observed heights at various time points.
### Complete Function
The resulting function based on our estimations could look like:
\[
h(t) = 200 - 5t^2
\]
where \( 5 \) is an adjusted value reflecting our observed changes against the temporal increments, postulating some minor air resistance in reality.
### Conclusion:
Thus, based on the observations and nature of motion equations, we have modeled the height of the object as \( h(t) = 200 - kt^2 \), where \( k \) is a proportionality constant linked to the effects of gravity and time. Further analysis and best fit techniques could refine this constant appropriately if precise data fitting is required.
### Data Overview
Here are the height measurements converted from feet to meters:
- The object starts at 200 feet, which is approximately 60.96 meters.
- The height decreases as time progresses, indicating a downward motion due to gravity.
### Choosing a Function
In free fall (neglecting air resistance), the height \( h(t) \) of an object at time \( t \) is generally modeled using the equations of motion under constant acceleration. The equation can be expressed as:
\[
h(t) = h_0 - \frac{1}{2} g t^2
\]
where:
- \( h_0 \) is the initial height,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) or about \( 10 \, \text{m/s}^2 \) for simplicity),
- \( t \) is the time in seconds.
### Conversion and Simplification
Given that the initial height \( h_0 \) is approximately \( 60.96 \) meters, we can represent the height function based on the data and we'll need to find a suitable value for \( g \).
### Fitting the Function
Using the data provided, let's initially establish whether a quadratic model is appropriate by observing the heights. The heights are decreasing in a somewhat quadratic manner, meaning that after 6 seconds, the decrease isn’t linear, which aligns well with a quadratic function.
We can refine the constant for gravity if we observe the given heights against the parameterized equation \( h(t) = 60.96 - \frac{1}{2} g t^2 \).
### Example Estimation
1. **From the Data:**
- At \( t = 0 \): \( h(0) = 200 \)
- At \( t = 1 \): \( h(1) \approx 195.1 \)
- At \( t = 2 \): \( h(2) \approx 180.4 \)
- At \( t = 3 \): \( h(3) \approx 155.9 \)
- At \( t = 4 \): \( h(4) \approx 121.6 \)
- At \( t = 5 \): \( h(5) \approx 77.5 \)
- At \( t = 6 \): \( h(6) \approx 23.6 \)
### Formulating the Function
To fit our data, we can perform a polynomial regression if necessary or simply derive constants through substitution and solving:
Assuming \( g = 9.81 \, \text{m/s}^2\) (or an average approximation similar), we can fit our model accordingly incorporating our observed heights at various time points.
### Complete Function
The resulting function based on our estimations could look like:
\[
h(t) = 200 - 5t^2
\]
where \( 5 \) is an adjusted value reflecting our observed changes against the temporal increments, postulating some minor air resistance in reality.
### Conclusion:
Thus, based on the observations and nature of motion equations, we have modeled the height of the object as \( h(t) = 200 - kt^2 \), where \( k \) is a proportionality constant linked to the effects of gravity and time. Further analysis and best fit techniques could refine this constant appropriately if precise data fitting is required.