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.
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.
Answers
GPT-4o mini
To model the scenario of the object falling from a cliff, we can use the basic physics of free fall under the influence of gravity. When an object is dropped (without any initial velocity) from a height, the distance it falls over time can be modeled by a quadratic function based on the kinematic equation:
\[ h(t) = h_0 - \frac{1}{2}gt^2 \]
where:
- \( h(t) \) is the height above ground at time \( t \),
- \( h_0 \) is the initial height (200 feet in this case),
- \( g \) is the acceleration due to gravity (approximately 32 ft/s² in the Imperial system),
- \( t \) is the time in seconds.
A. **Modeling the function:**
Given that the object starts from an initial height of 200 feet and falls under the influence of gravity, we can write:
\[ h(t) = 200 - 16t^2 \]
This formula was chosen because:
1. It starts at \( h(0) = 200 \) feet, which is correct for the initial condition.
2. The term \(-16t^2\) reflects the gravitational acceleration in feet per second squared (the factor of 16 comes from \(-\frac{1}{2}g\), where \(g = 32\) ft/s²).
B. **Graphing the function:**
For the graph, we use the function:
\[ h(t) = 200 - 16t^2 \]
Key features:
- The y-intercept is (0, 200).
- The object will hit the ground when \( h(t) = 0 \). Setting the equation to zero to find when that occurs:
\[ 200 - 16t^2 = 0 \]
\[ 16t^2 = 200 \]
\[ t^2 = 12.5 \]
\[ t = \sqrt{12.5} \approx 3.54 \text{ seconds} \]
The domain for the scenario is \( t \in [0, 3.54] \), where \( t \) is the time elapsed in seconds.
(This is a placeholder image. You would need to sketch or plot this in a graphing tool, showing the parabola decreasing from 200 to 0.)
C. **Finding the height at 3.5 seconds:**
To find the height at \( t = 3.5 \), substitute \( t = 3.5 \) into the height equation:
\[ h(3.5) = 200 - 16(3.5^2) \]
\[ h(3.5) = 200 - 16 \cdot 12.25 \]
\[ h(3.5) = 200 - 196 \]
\[ h(3.5) = 4 \text{ feet} \]
Therefore, the height of the object at 3.5 seconds is **4 feet**.
\[ h(t) = h_0 - \frac{1}{2}gt^2 \]
where:
- \( h(t) \) is the height above ground at time \( t \),
- \( h_0 \) is the initial height (200 feet in this case),
- \( g \) is the acceleration due to gravity (approximately 32 ft/s² in the Imperial system),
- \( t \) is the time in seconds.
A. **Modeling the function:**
Given that the object starts from an initial height of 200 feet and falls under the influence of gravity, we can write:
\[ h(t) = 200 - 16t^2 \]
This formula was chosen because:
1. It starts at \( h(0) = 200 \) feet, which is correct for the initial condition.
2. The term \(-16t^2\) reflects the gravitational acceleration in feet per second squared (the factor of 16 comes from \(-\frac{1}{2}g\), where \(g = 32\) ft/s²).
B. **Graphing the function:**
For the graph, we use the function:
\[ h(t) = 200 - 16t^2 \]
Key features:
- The y-intercept is (0, 200).
- The object will hit the ground when \( h(t) = 0 \). Setting the equation to zero to find when that occurs:
\[ 200 - 16t^2 = 0 \]
\[ 16t^2 = 200 \]
\[ t^2 = 12.5 \]
\[ t = \sqrt{12.5} \approx 3.54 \text{ seconds} \]
The domain for the scenario is \( t \in [0, 3.54] \), where \( t \) is the time elapsed in seconds.
(This is a placeholder image. You would need to sketch or plot this in a graphing tool, showing the parabola decreasing from 200 to 0.)
C. **Finding the height at 3.5 seconds:**
To find the height at \( t = 3.5 \), substitute \( t = 3.5 \) into the height equation:
\[ h(3.5) = 200 - 16(3.5^2) \]
\[ h(3.5) = 200 - 16 \cdot 12.25 \]
\[ h(3.5) = 200 - 196 \]
\[ h(3.5) = 4 \text{ feet} \]
Therefore, the height of the object at 3.5 seconds is **4 feet**.