if a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t second is given by y=65t-16t^2. find the average velocity for the time period beginning when t=2 and lasting 0.5 seconds, 0.1 seconds and 0.01 seconds


To find the average velocity of the ball over a given time period, we use the formula for average velocity:

\[
\text{Average Velocity} = \frac{\Delta y}{\Delta t}
\]

Where:
- \(\Delta y\) is the change in height (final height minus initial height)
- \(\Delta t\) is the change in time (final time minus initial time)

Given the height function:
\[
y(t) = 65t - 16t^2
\]

We will calculate the average velocity for three different time periods starting at \(t = 2\):

### Time period of 0.5 seconds:
The final time is \(t = 2 + 0.5 = 2.5\).

1. **Height at \(t = 2\)**:
   \[
   y(2) = 65(2) - 16(2)^2 = 130 - 64 = 66 \text{ feet}
   \]

2. **Height at \(t = 2.5\)**:
   \[
   y(2.5) = 65(2.5) - 16(2.5)^2 = 162.5 - 100 = 62.5 \text{ feet}
   \]

3. **Average velocity for \(t = 2\) to \(t = 2.5\)**:
   \[
   \text{Average Velocity} = \frac{62.5 - 66}{2.5 - 2} = \frac{-3.5}{0.5} = -7 \text{ ft/s}
   \]

### Time period of 0.1 seconds:
The final time is \(t = 2 + 0.1 = 2.1\).

1. **Height at \(t = 2.1\)**:
   \[
   y(2.1) = 65(2.1) - 16(2.1)^2 = 136.5 - 70.56 = 65.94 \text{ feet}
   \]

2. **Average velocity for \(t = 2\) to \(t = 2.1\)**:
   \[
   \text{Average Velocity} = \frac{65.94 - 66}{2.1 - 2} = \frac{-0.06}{0.1} = -0.6 \text{ ft/s}
   \]

### Time period of 0.01 seconds:
The final time is \(t = 2 + 0.01 = 2.01\).

1. **Height at \(t = 2.01\)**:
   \[
   y(2.01) = 65(2.01) - 16(2.01)^2 = 130.65 - 64.3216 = 66.3284 \text{ feet}
   \]

2. **Average velocity for \(t = 2\) to \(t = 2.01\)**:
   \[
   \text{Average Velocity} = \frac{66.3284 - 66}{2.01 - 2} = \frac{0.3284}{0.01} = 32.84 \text{ ft/s}
   \]

### Summary of Average Velocities:
- For 0.5 seconds: \(-7 \, \text{ft/s}\)
- For 0.1 seconds: \(-0.6 \, \text{ft/s}\)
- For 0.01 seconds: \(32.84 \, \text{ft/s}\)

Would you like more details on any of these calculations or clarifications?

Here are five related questions:
1. How would you calculate instantaneous velocity at \(t = 2\)?
2. What is the velocity function for this problem?
3. How does the average velocity change as \(\Delta t\) approaches zero?
4. What is the maximum height of the ball, and at what time does it occur?
5. What would the average velocity be over the time period from \(t = 1\) to \(t = 2\)?

**Tip:** The average velocity over smaller time intervals approaches the instantaneous velocity at a specific point.

If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t second is given by y=65t-16t^2. Find the average velocity for the time period beginning when t=2 and lasting 0.5 seconds, 0.1 seconds, and 0.01 seconds.

This problem involves finding the average velocity of a ball thrown straight up into the air with an initial velocity of 65 ft/s. It covers key concepts in kinematics and algebra, including quadratic equations and the difference quotient. The problem explores the average velocity over intervals of 0.5, 0.1, and 0.01 seconds, suitable for high school students in Grades 10-12.

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