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To solve this problem, let's first break it down.

The height \( y \) of the ball at time \( t \) is given by the equation:
\[
y = 45t - 16t^2
\]
### Part 1: Find the average velocity for different small time intervals.

The average velocity over a time interval \([t_1, t_2]\) can be found using the formula:
\[
\text{Average velocity} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}
\]
Here, \( t_1 = 2 \), and \( t_2 \) is determined by adding small increments to \( t_1 \) (0.1, 0.01, and 0.001 seconds respectively).

### (i) For \( t_2 = 2 + 0.1 = 2.1 \):
We need to calculate the average velocity between \( t = 2 \) and \( t = 2.1 \).

### (ii) For \( t_2 = 2 + 0.01 = 2.01 \):
We need to calculate the average velocity between \( t = 2 \) and \( t = 2.01 \).

### (iii) For \( t_2 = 2 + 0.001 = 2.001 \):
We need to calculate the average velocity between \( t = 2 \) and \( t = 2.001 \).

I'll now calculate these velocities step by step.The average velocities for the different time intervals are:

(i) For \( t_2 = 2.1 \) (time interval of 0.1 seconds): 
\[
\text{Average velocity} \approx -20.6 \, \text{ft/s}
\]

(ii) For \( t_2 = 2.01 \) (time interval of 0.01 seconds): 
\[
\text{Average velocity} \approx -19.16 \, \text{ft/s}
\]

(iii) For \( t_2 = 2.001 \) (time interval of 0.001 seconds): 
\[
\text{Average velocity} \approx -19.016 \, \text{ft/s}
\]

### Part 2: Instantaneous Velocity at \( t = 2 \)

The instantaneous velocity is the derivative of the height function, \( y = 45t - 16t^2 \), with respect to \( t \). The derivative of \( y \) is:

\[
v(t) = \frac{dy}{dt} = 45 - 32t
\]

We can now compute the instantaneous velocity at \( t = 2 \).The instantaneous velocity of the ball when \( t = 2 \) seconds is:

\[
\text{Instantaneous velocity} = -19 \, \text{ft/s}
\]

Would you like further details or clarifications?

Here are five related questions for further exploration:
1. How does the average velocity change as the time interval decreases?
2. What is the physical significance of negative velocity in this context?
3. Can you calculate the maximum height the ball reaches?
4. What is the time when the ball reaches the maximum height?
5. How would the problem change if the acceleration due to gravity were different?

**Tip:** The closer the average velocity over small intervals gets to the instantaneous velocity, the more precise your approximation becomes.

If a ball is thrown straight up into the air with an initial velocity of 45 ft/s, its height in feet after t seconds is given by y = 45t - 16t^2. Find the average velocity for the time period beginning when t = 2 and lasting (i) 0.1 seconds, (ii) 0.01 seconds, (iii) 0.001 seconds. Finally, based on the above results, guess what the instantaneous velocity of the ball is when t = 2.

Solve a problem involving a ball thrown straight up into the air with an initial velocity of 45 ft/s. Learn how to calculate the average velocity over different time intervals and estimate the instantaneous velocity using calculus and kinematics. Suitable for high school students.

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