Question
A small object is dropped from the top of a 200ft. 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.
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 height of the object falling from the cliff, we can use a quadratic function because the motion of a falling object under the influence of gravity is quadratic in nature. This type of motion can be represented by the standard equation of motion:
\[ h(t) = h_0 + v_0 t + \frac{1}{2} a t^2 \]
where:
- \( h(t) \) is the height at time \( t \),
- \( h_0 \) is the initial height (200 ft in this case),
- \( v_0 \) is the initial velocity (0 ft/s since the object is dropped),
- \( a \) is the acceleration due to gravity (approximately -32 ft/s² in feet).
Given that the object is released (not thrown), we can simplify our height function for immediate calculations:
\[ h(t) = 200 - 16t^2 \]
Here, we've converted the acceleration due to gravity into feet per second squared (-32 ft/s²) and used half of that as -16 for the quadratic term.
## A: Model the Scenario
1. **Choosing the Function**:
We used a quadratic function because the height of the object as a function of time follows the physics principles of free-fall motion, which is defined by a quadratic equation. The negative coefficient of \( t^2 \) reflects the downward acceleration due to gravity.
2. **Equation**:
The model we used was:
\[ h(t) = 200 - 16t^2 \]
## B: Sketch a Graph
To sketch the graph of this function:
1. Assign the x-axis to time (0 to 6 seconds).
2. Assign the y-axis to height (from the maximum of 200 ft to just above 0 ft).
Key features to highlight:
- Start point at \( (0, 200) \)
- The object will touch the ground at approximately the time it reaches 0 height, which we can determine by solving \( 200 - 16t^2 = 0 \):
- \( 16t^2 = 200 \)
- \( t^2 = 12.5 \)
- \( t = \sqrt{12.5} \approx 3.54 \) seconds
Thus, the object touches the ground roughly at \( t \approx 3.54 \) seconds.
The graph should show a downward-opening parabola starting at (0, 200) and curving down towards the x-axis, nearing (3.54, 0).
The domain of the scenario is \( t \in [0, 3.54] \).
## C: Height at 3.5 Seconds
To find the height at 3.5 seconds, we substitute \( t = 3.5 \) into our height function:
\[
h(3.5) = 200 - 16(3.5)^2
\]
Calculating it step-by-step:
\[
h(3.5) = 200 - 16 \times 12.25
\]
\[
h(3.5) = 200 - 196
\]
\[
h(3.5) = 4 \text{ ft}
\]
Thus, the height of the object at 3.5 seconds is **4 ft**.
\[ h(t) = h_0 + v_0 t + \frac{1}{2} a t^2 \]
where:
- \( h(t) \) is the height at time \( t \),
- \( h_0 \) is the initial height (200 ft in this case),
- \( v_0 \) is the initial velocity (0 ft/s since the object is dropped),
- \( a \) is the acceleration due to gravity (approximately -32 ft/s² in feet).
Given that the object is released (not thrown), we can simplify our height function for immediate calculations:
\[ h(t) = 200 - 16t^2 \]
Here, we've converted the acceleration due to gravity into feet per second squared (-32 ft/s²) and used half of that as -16 for the quadratic term.
## A: Model the Scenario
1. **Choosing the Function**:
We used a quadratic function because the height of the object as a function of time follows the physics principles of free-fall motion, which is defined by a quadratic equation. The negative coefficient of \( t^2 \) reflects the downward acceleration due to gravity.
2. **Equation**:
The model we used was:
\[ h(t) = 200 - 16t^2 \]
## B: Sketch a Graph
To sketch the graph of this function:
1. Assign the x-axis to time (0 to 6 seconds).
2. Assign the y-axis to height (from the maximum of 200 ft to just above 0 ft).
Key features to highlight:
- Start point at \( (0, 200) \)
- The object will touch the ground at approximately the time it reaches 0 height, which we can determine by solving \( 200 - 16t^2 = 0 \):
- \( 16t^2 = 200 \)
- \( t^2 = 12.5 \)
- \( t = \sqrt{12.5} \approx 3.54 \) seconds
Thus, the object touches the ground roughly at \( t \approx 3.54 \) seconds.
The graph should show a downward-opening parabola starting at (0, 200) and curving down towards the x-axis, nearing (3.54, 0).
The domain of the scenario is \( t \in [0, 3.54] \).
## C: Height at 3.5 Seconds
To find the height at 3.5 seconds, we substitute \( t = 3.5 \) into our height function:
\[
h(3.5) = 200 - 16(3.5)^2
\]
Calculating it step-by-step:
\[
h(3.5) = 200 - 16 \times 12.25
\]
\[
h(3.5) = 200 - 196
\]
\[
h(3.5) = 4 \text{ ft}
\]
Thus, the height of the object at 3.5 seconds is **4 ft**.