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
B. Sketch a graph that highlights the key features of the scenario. As you create your graph, consider the domain for this scenario.
C. What is the height of the object at 3.5 seconds? DO NOT ROUND.
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
B. Sketch a graph that highlights the key features of the scenario. As you create your graph, consider the domain for this scenario.
C. What is the height of the object at 3.5 seconds? DO NOT ROUND.
Answers
GPT-4o mini
### A. Model this scenario with a function
To model the height of an object dropped from a cliff, we can use quadratic modeling since the height of a free-falling object (neglecting air resistance) follows a parabolic trajectory.
The general function for the height \( h(t) \) of an object in free fall can generally be modeled by the equation:
\[
h(t) = -\frac{1}{2}gt^2 + h_0
\]
where:
- \( h(t) \) is the height at time \( t \),
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( h_0 \) is the initial height (in this case, \( 200 \, \text{ft} \)), which we need to convert to meters (\( 1 \, \text{ft} = 0.3048 \, \text{m} \), so \( 200 \, \text{ft} = 60.96 \, \text{m} \)).
Given that the acceleration due to gravity affects the height of the object negatively over time, we can write the height function as:
\[
h(t) = -\frac{1}{2}gt^2 + 60.96
\]
If our time intervals are in seconds, we will use \( g \approx 9.8 \, \text{m/s}^2 \). Thus, we would have:
\[
h(t) = -4.9t^2 + 60.96
\]
This function models the height of the object at any given second \( t \) after it has been dropped.
### B. Sketch a graph
#### Key features to include in the sketch:
- The initial height (when \( t = 0 \)) is \( 60.96 \, \text{m} \).
- The graph will have a downward-opening parabola since the object falls and the height decreases over time.
- For the domain, time \( t \) is non-negative (i.e., \( t \geq 0 \)). More practically, in this scenario, the time can be limited to around \( 6 \) seconds since we have data for that timeframe and the object hits the ground before then.
The x-axis will represent time (from \( 0 \) to \( 6 \) seconds), and the y-axis will represent the height (from \( 0 \) to \( 60.96 \, \text{m} \)).
(Unfortunately, I can't provide a graphical sketch directly here, but you would plot \( t \) on the x-axis from \( 0 \) to \( 6 \) and \( h(t) \) on the y-axis up to \( 60.96 \). You can calculate a few more points using the provided function for extra data points, or simply plot the given data points to represent the descent.)
### C. What is the height of the object at 3.5 seconds?
To find the height at \( t = 3.5 \) seconds, plug \( 3.5 \) into the function we established:
\[
h(3.5) = -4.9(3.5)^2 + 60.96
\]
Calculating it step by step:
1. Calculate \( (3.5)^2 = 12.25 \)
2. Now calculate \( -4.9 \cdot 12.25 \):
\[
-4.9 \cdot 12.25 = -60.025
\]
3. Finally, add \( 60.96 \):
\[
h(3.5) = -60.025 + 60.96 = 0.935
\]
So, the height of the object at \( 3.5 \) seconds is:
\[
\text{Height} = 0.935 \, \text{meters}
\]
To model the height of an object dropped from a cliff, we can use quadratic modeling since the height of a free-falling object (neglecting air resistance) follows a parabolic trajectory.
The general function for the height \( h(t) \) of an object in free fall can generally be modeled by the equation:
\[
h(t) = -\frac{1}{2}gt^2 + h_0
\]
where:
- \( h(t) \) is the height at time \( t \),
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( h_0 \) is the initial height (in this case, \( 200 \, \text{ft} \)), which we need to convert to meters (\( 1 \, \text{ft} = 0.3048 \, \text{m} \), so \( 200 \, \text{ft} = 60.96 \, \text{m} \)).
Given that the acceleration due to gravity affects the height of the object negatively over time, we can write the height function as:
\[
h(t) = -\frac{1}{2}gt^2 + 60.96
\]
If our time intervals are in seconds, we will use \( g \approx 9.8 \, \text{m/s}^2 \). Thus, we would have:
\[
h(t) = -4.9t^2 + 60.96
\]
This function models the height of the object at any given second \( t \) after it has been dropped.
### B. Sketch a graph
#### Key features to include in the sketch:
- The initial height (when \( t = 0 \)) is \( 60.96 \, \text{m} \).
- The graph will have a downward-opening parabola since the object falls and the height decreases over time.
- For the domain, time \( t \) is non-negative (i.e., \( t \geq 0 \)). More practically, in this scenario, the time can be limited to around \( 6 \) seconds since we have data for that timeframe and the object hits the ground before then.
The x-axis will represent time (from \( 0 \) to \( 6 \) seconds), and the y-axis will represent the height (from \( 0 \) to \( 60.96 \, \text{m} \)).
(Unfortunately, I can't provide a graphical sketch directly here, but you would plot \( t \) on the x-axis from \( 0 \) to \( 6 \) and \( h(t) \) on the y-axis up to \( 60.96 \). You can calculate a few more points using the provided function for extra data points, or simply plot the given data points to represent the descent.)
### C. What is the height of the object at 3.5 seconds?
To find the height at \( t = 3.5 \) seconds, plug \( 3.5 \) into the function we established:
\[
h(3.5) = -4.9(3.5)^2 + 60.96
\]
Calculating it step by step:
1. Calculate \( (3.5)^2 = 12.25 \)
2. Now calculate \( -4.9 \cdot 12.25 \):
\[
-4.9 \cdot 12.25 = -60.025
\]
3. Finally, add \( 60.96 \):
\[
h(3.5) = -60.025 + 60.96 = 0.935
\]
So, the height of the object at \( 3.5 \) seconds is:
\[
\text{Height} = 0.935 \, \text{meters}
\]