Question
A ball is thrown off a low cliff to the beach below. It is thrown into the air with an initial upward velocity of 5 meters per second. The following graph models its height above sea level in meters over time in seconds. Would a linear function, a quadratic function, or an exponential function be most appropriate as a model for this graph? Explain your choice. already done.
The best way to model the height of the ball is with a quadratic function. This is because when you throw something up, it goes up to a peak and then comes back down, which makes a U-shaped graph. A linear function would look like a straight line, and an exponential function goes up really fast, like a rocket, which isn't what happens here. So, a quadratic function shows how the ball goes up, stops, and then falls back down, just like the graph.
Use your table to create an equation that models this graph and scenario.
The best way to model the height of the ball is with a quadratic function. This is because when you throw something up, it goes up to a peak and then comes back down, which makes a U-shaped graph. A linear function would look like a straight line, and an exponential function goes up really fast, like a rocket, which isn't what happens here. So, a quadratic function shows how the ball goes up, stops, and then falls back down, just like the graph.
Use your table to create an equation that models this graph and scenario.
Answers
GPT-4o mini
To create a quadratic model for the height of the ball over time, we will use the standard form of a quadratic equation:
\[
h(t) = at^2 + bt + c
\]
where:
- \(h(t)\) is the height of the ball at time \(t\),
- \(a\), \(b\), and \(c\) are constants.
Given that the ball is thrown with an initial upward velocity of \(5\) meters per second from a low cliff, we can assume:
1. The initial height \(c\) is at the height of the cliff (let’s denote this as \(h_0\)).
2. The initial velocity will influence the linear term \(b\).
3. The acceleration due to gravity \(a\) is approximately \(-9.81\) meters per second squared (which will be a negative value indicating that it is acting downwards).
To construct the equation, we need to define the parameters based on the scenario:
1. **Initial height (\(h_0\))**: Let's assume the cliff is at \(h_0\) meters above sea level. If you don't have this value, you can denote it as \(c\).
2. **Initial velocity (\(v_0 = 5\) m/s)**: This will be used for the derivative at \(t = 0\).
3. **Acceleration due to gravity**: We will use \(-4.905\) for \(a\) since this is half of \(-9.81\) due to the formula being in terms of seconds squared.
So, we can summarize:
- \(a = -4.905\) (half of the acceleration due to gravity),
- \(b = 5\) (the initial upward velocity),
- \(c = h_0\) (initial height of the cliff).
Putting this all together, the model for height \(h(t)\) can be expressed as:
\[
h(t) = -4.905t^2 + 5t + h_0
\]
### Example with Values
If we assume the height of the cliff \(h_0\) is, for example, \(10\) meters, then our specific equation becomes:
\[
h(t) = -4.905t^2 + 5t + 10
\]
This quadratic function effectively models the height of the ball over time, showing its upward motion and eventual descent back down to sea level. You can adjust \(h_0\) according to the actual height of the cliff if needed.
\[
h(t) = at^2 + bt + c
\]
where:
- \(h(t)\) is the height of the ball at time \(t\),
- \(a\), \(b\), and \(c\) are constants.
Given that the ball is thrown with an initial upward velocity of \(5\) meters per second from a low cliff, we can assume:
1. The initial height \(c\) is at the height of the cliff (let’s denote this as \(h_0\)).
2. The initial velocity will influence the linear term \(b\).
3. The acceleration due to gravity \(a\) is approximately \(-9.81\) meters per second squared (which will be a negative value indicating that it is acting downwards).
To construct the equation, we need to define the parameters based on the scenario:
1. **Initial height (\(h_0\))**: Let's assume the cliff is at \(h_0\) meters above sea level. If you don't have this value, you can denote it as \(c\).
2. **Initial velocity (\(v_0 = 5\) m/s)**: This will be used for the derivative at \(t = 0\).
3. **Acceleration due to gravity**: We will use \(-4.905\) for \(a\) since this is half of \(-9.81\) due to the formula being in terms of seconds squared.
So, we can summarize:
- \(a = -4.905\) (half of the acceleration due to gravity),
- \(b = 5\) (the initial upward velocity),
- \(c = h_0\) (initial height of the cliff).
Putting this all together, the model for height \(h(t)\) can be expressed as:
\[
h(t) = -4.905t^2 + 5t + h_0
\]
### Example with Values
If we assume the height of the cliff \(h_0\) is, for example, \(10\) meters, then our specific equation becomes:
\[
h(t) = -4.905t^2 + 5t + 10
\]
This quadratic function effectively models the height of the ball over time, showing its upward motion and eventual descent back down to sea level. You can adjust \(h_0\) according to the actual height of the cliff if needed.